Class 12: Tanget and Normals

$\displaystyle \textbf{Question 1: }~\text{Find the equation of the tangent to the curve }\sqrt{x}+\sqrt{y}=a, \\ \text{ at the point }\left(\frac{a^2}{4},\frac{a^2}{4}\right).$
$\displaystyle \text{Answer:}$

$\displaystyle \sqrt{x} + \sqrt{y} = a$

$\displaystyle \text{Differentiating both sides with respect to } x,$

$\displaystyle \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0$

$\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}$

$\displaystyle \text{Given point } (x_1, y_1) = \left(\frac{a^2}{4}, \frac{a^2}{4}\right)$

$\displaystyle \text{Slope of tangent, } m = \left.\frac{dy}{dx}\right|_{\left(\frac{a^2}{4}, \frac{a^2}{4}\right)} = -\frac{\sqrt{\frac{a^2}{4}}}{\sqrt{\frac{a^2}{4}}} = -1$

$\displaystyle \text{Equation of tangent is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y - \frac{a^2}{4} = -1\left(x - \frac{a^2}{4}\right)$

$\displaystyle \Rightarrow y - \frac{a^2}{4} = -x + \frac{a^2}{4}$

$\displaystyle \Rightarrow x + y = \frac{a^2}{2}$

$\displaystyle \textbf{Question 2: }~\text{Find the equation of the normal to }y=2x^3-x^2+3\text{ at }(1,4).$
$\displaystyle \text{Answer:}$

$\displaystyle y = 2x^3 - x^2 + 3$

$\displaystyle \text{Differentiating both sides with respect to } x,$

$\displaystyle \frac{dy}{dx} = 6x^2 - 2x$

$\displaystyle \text{Slope of tangent at } (1,4) = \left.\frac{dy}{dx}\right|_{x=1} = 6(1)^2 - 2(1) = 4$

$\displaystyle \text{Slope of normal } = -\frac{1}{\text{Slope of tangent}} = -\frac{1}{4}$

$\displaystyle \text{Given point } (x_1, y_1) = (1,4)$

$\displaystyle \text{Equation of normal is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y - 4 = -\frac{1}{4}(x - 1)$

$\displaystyle \Rightarrow 4y - 16 = -x + 1$

$\displaystyle \Rightarrow x + 4y = 17$

$\displaystyle \textbf{Question 3: }~\text{Find the equations of the tangent and the normal to the following curves} \\ \text{at the indicated points:}$
$\displaystyle \text{(i) }\;y=x^4-bx^3+13x^2-10x+5\text{ at }(0,5)$

$\displaystyle \text{Answer:}$

$\displaystyle y = x^4 - bx^3 + 13x^2 - 10x + 5$

$\displaystyle \text{Differentiating both sides with respect to } x,$

$\displaystyle \frac{dy}{dx} = 4x^3 - 3bx^2 + 26x - 10$

$\displaystyle \text{Slope of tangent at } (0,5) = \left.\frac{dy}{dx}\right|_{x=0} = 4(0)^3 - 3b(0)^2 + 26(0) - 10 = -10$

$\displaystyle \text{Given point } (x_1, y_1) = (0,5)$

$\displaystyle \text{Equation of tangent is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y - 5 = -10(x - 0)$

$\displaystyle \Rightarrow y - 5 = -10x$

$\displaystyle \Rightarrow y + 10x - 5 = 0$

$\displaystyle \text{Equation of normal is } y - y_1 = -\frac{1}{m}(x - x_1)$

$\displaystyle \Rightarrow y - 5 = \frac{1}{10}(x - 0)$

$\displaystyle \Rightarrow 10y - 50 = x$

$\displaystyle \Rightarrow x - 10y + 50 = 0$

$\displaystyle \text{(ii) }\;y=x^4-6x^3+13x^2-10x+5\text{ at }x=1\;[\text{ CBSE 2011}]$

$\displaystyle \text{Answer:}$

$\displaystyle y = x^4 - 6x^3 + 13x^2 - 10x + 5$

$\displaystyle \text{When } x = 1,$

$\displaystyle y = 1 - 6 + 13 - 10 + 5 = 3$

$\displaystyle \text{So, } (x_1, y_1) = (1,3)$

$\displaystyle \text{Differentiating both sides with respect to } x,$

$\displaystyle \frac{dy}{dx} = 4x^3 - 18x^2 + 26x - 10$

$\displaystyle \text{Slope of tangent at } (1,3) = \left.\frac{dy}{dx}\right|_{x=1} = 4 - 18 + 26 - 10 = 2$

$\displaystyle \text{Equation of tangent is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y - 3 = 2(x - 1)$

$\displaystyle \Rightarrow y - 3 = 2x - 2$

$\displaystyle \Rightarrow 2x - y + 1 = 0$

$\displaystyle \text{Equation of normal is } y - y_1 = -\frac{1}{m}(x - x_1)$

$\displaystyle \Rightarrow y - 3 = -\frac{1}{2}(x - 1)$

$\displaystyle \Rightarrow 2y - 6 = -x + 1$

$\displaystyle \Rightarrow x + 2y - 7 = 0$

$\displaystyle \text{(iii) }\;y=x^2\text{ at }(0,0)$

$\displaystyle \text{Answer:}$

$\displaystyle y = x^2$

$\displaystyle \text{Differentiating both sides with respect to } x,$

$\displaystyle \frac{dy}{dx} = 2x$

$\displaystyle \text{Given point } (x_1, y_1) = (0,0)$

$\displaystyle \text{Slope of tangent at } (0,0) = \left.\frac{dy}{dx}\right|_{x=0} = 2(0) = 0$

$\displaystyle \text{Equation of tangent is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y - 0 = 0(x - 0)$

$\displaystyle \Rightarrow y = 0$

$\displaystyle \text{Slope of normal is undefined since } m = 0$

$\displaystyle \text{Hence, the equation of the normal is } x = 0$

$\displaystyle \text{(iv) }\;y=2x^2-3x-1\text{ at }(1,-2)$

$\displaystyle \text{Answer:}$

$\displaystyle y = 2x^2 - 3x - 1$

$\displaystyle \text{Differentiating both sides with respect to } x,$

$\displaystyle \frac{dy}{dx} = 4x - 3$

$\displaystyle \text{Given point } (x_1, y_1) = (1,-2)$

$\displaystyle \text{Slope of tangent at } (1,-2) = \left.\frac{dy}{dx}\right|_{x=1} = 4(1) - 3 = 1$

$\displaystyle \text{Equation of tangent is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y + 2 = 1(x - 1)$

$\displaystyle \Rightarrow y + 2 = x - 1$

$\displaystyle \Rightarrow x - y - 3 = 0$

$\displaystyle \text{Equation of normal is } y - y_1 = -\frac{1}{m}(x - x_1)$

$\displaystyle \Rightarrow y + 2 = -1(x - 1)$

$\displaystyle \Rightarrow y + 2 = -x + 1$

$\displaystyle \Rightarrow x + y + 1 = 0$

$\displaystyle \text{(v) }\;y^2=\frac{x^3}{4-x}\text{ at }(2,-2)$

$\displaystyle \text{Answer:}$

$\displaystyle y^2 = \frac{x^3}{4 - x}$

$\displaystyle \text{Differentiating both sides with respect to } x,$

$\displaystyle 2y \frac{dy}{dx} = \frac{(4 - x)(3x^2) - x^3(-1)}{(4 - x)^2}$

$\displaystyle 2y \frac{dy}{dx} = \frac{12x^2 - 3x^3 + x^3}{(4 - x)^2} = \frac{12x^2 - 2x^3}{(4 - x)^2}$

$\displaystyle \frac{dy}{dx} = \frac{12x^2 - 2x^3}{2y(4 - x)^2}$

$\displaystyle \text{Given point } (x_1, y_1) = (2,-2)$

$\displaystyle \text{Slope of tangent at } (2,-2) = \left.\frac{dy}{dx}\right|_{(2,-2)} = \frac{12(2)^2 - 2(2)^3}{2(-2)(4 - 2)^2} = \frac{48 - 16}{-16} = -2$

$\displaystyle \text{Equation of tangent is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y + 2 = -2(x - 2)$

$\displaystyle \Rightarrow y + 2 = -2x + 4$

$\displaystyle \Rightarrow 2x + y - 2 = 0$

$\displaystyle \text{Equation of normal is } y - y_1 = -\frac{1}{m}(x - x_1)$

$\displaystyle \Rightarrow y + 2 = \frac{1}{2}(x - 2)$

$\displaystyle \Rightarrow 2y + 4 = x - 2$

$\displaystyle \Rightarrow x - 2y - 6 = 0$

$\displaystyle \text{(vi) }\;y=x^2+4x+1\text{ at }x=3\;[\text{CBSE 2004}]$

$\displaystyle \text{Answer:}$

$\displaystyle y = x^2 + 4x + 1$

$\displaystyle \text{Differentiating both sides with respect to } x,$

$\displaystyle \frac{dy}{dx} = 2x + 4$

$\displaystyle \text{When } x = 3,$

$\displaystyle y = 3^2 + 4(3) + 1 = 9 + 12 + 1 = 22$

$\displaystyle \text{So, } (x_1, y_1) = (3,22)$

$\displaystyle \text{Slope of tangent at } x = 3 = \left.\frac{dy}{dx}\right|_{x=3} = 2(3) + 4 = 10$

$\displaystyle \text{Equation of tangent is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y - 22 = 10(x - 3)$

$\displaystyle \Rightarrow y - 22 = 10x - 30$

$\displaystyle \Rightarrow 10x - y - 8 = 0$

$\displaystyle \text{Equation of normal is } y - y_1 = -\frac{1}{m}(x - x_1)$

$\displaystyle \Rightarrow y - 22 = -\frac{1}{10}(x - 3)$

$\displaystyle \Rightarrow 10y - 220 = -x + 3$

$\displaystyle \Rightarrow x + 10y - 223 = 0$

$\displaystyle \text{(vii) }\;\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{ at }(a\cos\theta,\;b\sin\theta)$

$\displaystyle \text{Answer:}$

$\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

$\displaystyle \text{Differentiating both sides with respect to } x,$

$\displaystyle \Rightarrow \frac{2x}{a^2} + \frac{2y}{b^2}\frac{dy}{dx} = 0$

$\displaystyle \Rightarrow \frac{2y}{b^2}\frac{dy}{dx} = -\frac{2x}{a^2}$

$\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{xb^2}{ya^2}$

$\displaystyle \text{Given point } (x_1, y_1) = (a\cos\theta, b\sin\theta)$

$\displaystyle \text{Slope of tangent at } (a\cos\theta, b\sin\theta) = \left.\frac{dy}{dx}\right|_{(a\cos\theta, b\sin\theta)} = -\frac{a\cos\theta\, b^2}{b\sin\theta\, a^2} = -\frac{b\cos\theta}{a\sin\theta}$

$\displaystyle \text{Equation of tangent is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y - b\sin\theta = -\frac{b\cos\theta}{a\sin\theta}(x - a\cos\theta)$

$\displaystyle \Rightarrow ay\sin\theta - ab\sin^2\theta = -bx\cos\theta + ab\cos^2\theta$

$\displaystyle \Rightarrow bx\cos\theta + ay\sin\theta = ab$

$\displaystyle \Rightarrow \frac{x}{a}\cos\theta + \frac{y}{b}\sin\theta = 1$

$\displaystyle \text{Equation of normal is } y - y_1 = -\frac{1}{m}(x - x_1)$

$\displaystyle \Rightarrow y - b\sin\theta = \frac{a\sin\theta}{b\cos\theta}(x - a\cos\theta)$

$\displaystyle \Rightarrow by\cos\theta - b^2\sin\theta\cos\theta = ax\sin\theta - a^2\sin\theta\cos\theta$

$\displaystyle \Rightarrow ax\sin\theta - by\cos\theta = (a^2 - b^2)\sin\theta\cos\theta$

$\displaystyle \Rightarrow ax\sec\theta - by\,\mathrm{cosec}\,\theta = a^2 - b^2$

$\displaystyle \text{(viii) }\;\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\text{ at }(a\sec\theta,\;b\tan\theta)$

$\displaystyle \text{Answer:}$

$\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

$\displaystyle \text{Differentiating both sides with respect to } x,$

$\displaystyle \Rightarrow \frac{2x}{a^2} - \frac{2y}{b^2}\frac{dy}{dx} = 0$

$\displaystyle \Rightarrow \frac{2y}{b^2}\frac{dy}{dx} = \frac{2x}{a^2}$

$\displaystyle \Rightarrow \frac{dy}{dx} = \frac{xb^2}{ya^2}$

$\displaystyle \text{Given point } (x_1, y_1) = (a\sec\theta, b\tan\theta)$

$\displaystyle \text{Slope of tangent at } (a\sec\theta, b\tan\theta) = \left.\frac{dy}{dx}\right|_{(a\sec\theta, b\tan\theta)} = \frac{a\sec\theta \cdot b^2}{b\tan\theta \cdot a^2} = \frac{b}{a\sin\theta}$

$\displaystyle \text{Equation of tangent is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y - b\tan\theta = \frac{b}{a\sin\theta}(x - a\sec\theta)$

$\displaystyle \Rightarrow ay\sin\theta - ab\sin\theta\tan\theta = bx - ab\sec\theta$

$\displaystyle \Rightarrow ay\sin\theta\cos\theta - ab\sin^2\theta = bx\cos\theta - ab$

$\displaystyle \Rightarrow bx\cos\theta - ay\sin\theta\cos\theta = ab(1 - \sin^2\theta)$

$\displaystyle \Rightarrow bx\cos\theta - ay\sin\theta\cos\theta = ab\cos^2\theta$

$\displaystyle \Rightarrow \frac{x}{a}\sec\theta - \frac{y}{b}\tan\theta = 1$

$\displaystyle \text{Equation of normal is } y - y_1 = -\frac{1}{m}(x - x_1)$

$\displaystyle \Rightarrow y - b\tan\theta = -\frac{a\sin\theta}{b}(x - a\sec\theta)$

$\displaystyle \Rightarrow yb - b^2\tan\theta = -ax\sin\theta + a^2\tan\theta$

$\displaystyle \Rightarrow ax\sin\theta + by = (a^2 + b^2)\tan\theta$

$\displaystyle \Rightarrow ax\cos\theta + by\,\mathrm{cot}\,\theta = a^2 + b^2$

$\displaystyle \text{(ix) }\;y^2=4ax\text{ at }\left(\frac{a}{m^2},\frac{2a}{m}\right)$

$\displaystyle \text{Answer:}$

$\displaystyle y^2 = 4ax$

$\displaystyle \text{Differentiating both sides with respect to } x,$

$\displaystyle 2y\frac{dy}{dx} = 4a$

$\displaystyle \Rightarrow \frac{dy}{dx} = \frac{2a}{y}$

$\displaystyle \text{Given point } (x_1, y_1) = \left(\frac{a}{m^2}, \frac{2a}{m}\right)$

$\displaystyle \text{Slope of tangent at } \left(\frac{a}{m^2}, \frac{2a}{m}\right) = \left.\frac{dy}{dx}\right|_{\left(\frac{a}{m^2}, \frac{2a}{m}\right)} = \frac{2a}{\frac{2a}{m}} = m$

$\displaystyle \text{Equation of tangent is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y - \frac{2a}{m} = m\left(x - \frac{a}{m^2}\right)$

$\displaystyle \Rightarrow my - 2a = m^2x - a$

$\displaystyle \Rightarrow m^2x - my + a = 0$

$\displaystyle \text{Equation of normal is } y - y_1 = -\frac{1}{m}(x - x_1)$

$\displaystyle \Rightarrow y - \frac{2a}{m} = -\frac{1}{m}\left(x - \frac{a}{m^2}\right)$

$\displaystyle \Rightarrow my - 2a = -\frac{1}{m}(m^2x - a)$

$\displaystyle \Rightarrow m^2y - 2am = -m^2x + a$

$\displaystyle \Rightarrow m^2x + m^2y - 2am - a = 0$

$\displaystyle \text{(x) }\;c^2(x^2+y^2)=x^2y^2\text{ at }\left(\frac{c}{\cos\theta},\frac{c}{\sin\theta}\right)$

$\displaystyle \text{Answer:}$

$\displaystyle c^2(x^2 + y^2) = x^2y^2$

$\displaystyle \text{Differentiating both sides with respect to } x,$

$\displaystyle \Rightarrow 2c^2x + 2c^2y\frac{dy}{dx} = 2xy^2 + 2x^2y\frac{dy}{dx}$

$\displaystyle \Rightarrow \frac{dy}{dx}(2c^2y - 2x^2y) = 2xy^2 - 2c^2x$

$\displaystyle \Rightarrow \frac{dy}{dx} = \frac{xy^2 - xc^2}{yc^2 - x^2y}$

$\displaystyle \text{Given point } (x_1, y_1) = \left(\frac{c}{\cos\theta}, \frac{c}{\sin\theta}\right)$

$\displaystyle \text{Slope of tangent at } \left(\frac{c}{\cos\theta}, \frac{c}{\sin\theta}\right) = \left.\frac{dy}{dx}\right|_{\left(\frac{c}{\cos\theta}, \frac{c}{\sin\theta}\right)}$

$\displaystyle = \frac{\frac{c}{\cos\theta}\left(\frac{c^2}{\sin^2\theta} - c^2\right)} {\frac{c}{\sin\theta}\left(c^2 - \frac{c^2}{\cos^2\theta}\right)}$

$\displaystyle = \frac{\cos^2\theta - 1}{1 - \sin^2\theta} \cdot \frac{\cos^3\theta}{\sin^3\theta} = -\frac{\cos^3\theta}{\sin^3\theta}$

$\displaystyle \text{Equation of tangent is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y - \frac{c}{\sin\theta} = -\frac{\cos^3\theta}{\sin^3\theta} \left(x - \frac{c}{\cos\theta}\right)$

$\displaystyle \Rightarrow \sin^2\theta(y\sin\theta - c) = -\cos^2\theta(x\cos\theta - c)$

$\displaystyle \Rightarrow x\cos^3\theta + y\sin^3\theta = c(\cos^2\theta + \sin^2\theta)$

$\displaystyle \Rightarrow x\cos^3\theta + y\sin^3\theta = c$

$\displaystyle \text{Equation of normal is } y - y_1 = -\frac{1}{m}(x - x_1)$

$\displaystyle \Rightarrow y - \frac{c}{\sin\theta} = \frac{\sin^3\theta}{\cos^3\theta} \left(x - \frac{c}{\cos\theta}\right)$

$\displaystyle \Rightarrow x\sin^3\theta - y\cos^3\theta = c\left(\frac{\sin^4\theta - \cos^4\theta}{\sin\theta\cos\theta}\right)$

$\displaystyle \Rightarrow x\sin^3\theta - y\cos^3\theta = c\left(\frac{(\sin^2\theta + \cos^2\theta) (\sin^2\theta - \cos^2\theta)}{\sin\theta\cos\theta}\right)$

$\displaystyle \Rightarrow x\sin^3\theta - y\cos^3\theta = c\left(\frac{-\cos 2\theta}{\sin\theta\cos\theta}\right)$

$\displaystyle \Rightarrow x\sin^3\theta - y\cos^3\theta = -2c\,\mathrm{cot}\,2\theta$

$\displaystyle \Rightarrow x\sin^3\theta - y\cos^3\theta + 2c\,\mathrm{cot}\,2\theta = 0$

$\displaystyle \text{(xi) }\;xy=c^2\text{ at }\left(ct,\frac{c}{t}\right)$

$\displaystyle \text{Answer:}$

$\displaystyle xy = c^2$

$\displaystyle \text{Differentiating both sides with respect to } x,$

$\displaystyle x\frac{dy}{dx} + y = 0$

$\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$

$\displaystyle \text{Given point } (x_1, y_1) = (ct, \tfrac{c}{t})$

$\displaystyle \text{Slope of tangent at } (ct, \tfrac{c}{t}) = \left.\frac{dy}{dx}\right|_{(ct,\,c/t)} = -\frac{c/t}{ct} = -\frac{1}{t^2}$

$\displaystyle \text{Equation of tangent is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y - \frac{c}{t} = -\frac{1}{t^2}(x - ct)$

$\displaystyle \Rightarrow y = -\frac{x}{t^2} + \frac{2c}{t}$

$\displaystyle \Rightarrow x + t^2y - 2ct = 0$

$\displaystyle \text{Equation of normal is } y - y_1 = -\frac{1}{m}(x - x_1)$

$\displaystyle \Rightarrow y - \frac{c}{t} = t^2(x - ct)$

$\displaystyle \Rightarrow y = t^2x - ct^3 + \frac{c}{t}$

$\displaystyle \Rightarrow t^3x - ty - ct^4 + c = 0$

$\displaystyle \text{(xii) }\;\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{ at }(x_1,y_1)$

$\displaystyle \text{Answer:}$

$\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

$\displaystyle \text{Differentiating both sides with respect to } x,$

$\displaystyle \Rightarrow \frac{2x}{a^2} + \frac{2y}{b^2}\frac{dy}{dx} = 0$

$\displaystyle \Rightarrow \frac{2y}{b^2}\frac{dy}{dx} = -\frac{2x}{a^2}$

$\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{xb^2}{ya^2}$

$\displaystyle \text{Given point } (x_1, y_1)$

$\displaystyle \text{Slope of tangent at } (x_1, y_1) = \left.\frac{dy}{dx}\right|_{(x_1,y_1)} = -\frac{x_1 b^2}{y_1 a^2}$

$\displaystyle \text{Equation of tangent is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y - y_1 = -\frac{x_1 b^2}{y_1 a^2}(x - x_1)$

$\displaystyle \Rightarrow yy_1a^2 - y_1^2a^2 = -xx_1b^2 + x_1^2b^2$

$\displaystyle \Rightarrow xx_1b^2 + yy_1a^2 = x_1^2b^2 + y_1^2a^2 \quad (1)$

$\displaystyle \text{Since } (x_1, y_1) \text{ lies on the curve,}$

$\displaystyle \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} = 1$

$\displaystyle \Rightarrow x_1^2b^2 + y_1^2a^2 = a^2b^2$

$\displaystyle \text{Substituting in (1),}$

$\displaystyle xx_1b^2 + yy_1a^2 = a^2b^2$

$\displaystyle \Rightarrow \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$

$\displaystyle \text{Equation of normal is } y - y_1 = -\frac{1}{m}(x - x_1)$

$\displaystyle \Rightarrow y - y_1 = \frac{y_1a^2}{x_1b^2}(x - x_1)$

$\displaystyle \Rightarrow yx_1b^2 - y_1x_1b^2 = xy_1a^2 - x_1y_1a^2$

$\displaystyle \Rightarrow xy_1a^2 - yx_1b^2 = x_1y_1(a^2 - b^2)$

$\displaystyle \Rightarrow \frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$

$\displaystyle \text{(xiii) }\;\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\text{ at }(x_0,y_0)$

$\displaystyle \text{Answer:}$

$\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

$\displaystyle \text{Differentiating both sides with respect to } x,$

$\displaystyle \Rightarrow \frac{2x}{a^2} - \frac{2y}{b^2}\frac{dy}{dx} = 0$

$\displaystyle \Rightarrow \frac{2y}{b^2}\frac{dy}{dx} = \frac{2x}{a^2}$

$\displaystyle \Rightarrow \frac{dy}{dx} = \frac{xb^2}{ya^2}$

$\displaystyle \text{Given point } (x_0, y_0)$

$\displaystyle \text{Slope of tangent at } (x_0, y_0) = \left.\frac{dy}{dx}\right|_{(x_0,y_0)} = \frac{x_0 b^2}{y_0 a^2}$

$\displaystyle \text{Equation of tangent is } y - y_0 = m(x - x_0)$

$\displaystyle \Rightarrow y - y_0 = \frac{x_0 b^2}{y_0 a^2}(x - x_0)$

$\displaystyle \Rightarrow yy_0a^2 - y_0^2a^2 = xx_0b^2 - x_0^2b^2$

$\displaystyle \Rightarrow xx_0b^2 - yy_0a^2 = x_0^2b^2 - y_0^2a^2 \quad (1)$

$\displaystyle \text{Since } (x_0, y_0) \text{ lies on the curve,}$

$\displaystyle \frac{x_0^2}{a^2} - \frac{y_0^2}{b^2} = 1$

$\displaystyle \Rightarrow x_0^2b^2 - y_0^2a^2 = a^2b^2$

$\displaystyle \text{Substituting in (1),}$

$\displaystyle xx_0b^2 - yy_0a^2 = a^2b^2$

$\displaystyle \Rightarrow \frac{xx_0}{a^2} - \frac{yy_0}{b^2} = 1$

$\displaystyle \text{Equation of normal is } y - y_0 = -\frac{1}{m}(x - x_0)$

$\displaystyle \Rightarrow y - y_0 = -\frac{y_0a^2}{x_0b^2}(x - x_0)$

$\displaystyle \Rightarrow yx_0b^2 - y_0x_0b^2 = -xy_0a^2 + x_0y_0a^2$

$\displaystyle \Rightarrow xy_0a^2 + yx_0b^2 = x_0y_0(a^2 + b^2)$

$\displaystyle \Rightarrow \frac{a^2x}{x_0} + \frac{b^2y}{y_0} = a^2 + b^2$

$\displaystyle \text{(xiv) }\;x^{2/3}+y^{2/3}=2\text{ at }(1,1)$

$\displaystyle \text{Answer:}$

$\displaystyle x^{\frac{2}{3}} + y^{\frac{2}{3}} = 2$

$\displaystyle \text{Differentiating both sides with respect to } x,$

$\displaystyle \frac{2}{3}x^{-\frac{1}{3}} + \frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx} = 0$

$\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}} = -\frac{y^{\frac{1}{3}}}{x^{\frac{1}{3}}}$

$\displaystyle \text{Given point } (x_1, y_1) = (1,1)$

$\displaystyle \text{Slope of tangent at } (1,1) = \left.\frac{dy}{dx}\right|_{(1,1)} = -\frac{1}{1} = -1$

$\displaystyle \text{Equation of tangent is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y - 1 = -1(x - 1)$

$\displaystyle \Rightarrow y - 1 = -x + 1$

$\displaystyle \Rightarrow x + y - 2 = 0$

$\displaystyle \text{Equation of normal is } y - y_1 = -\frac{1}{m}(x - x_1)$

$\displaystyle \Rightarrow y - 1 = 1(x - 1)$

$\displaystyle \Rightarrow y - x = 0$

$\displaystyle \text{(xv) }\;x^2=4y\text{ at }(2,1)$

$\displaystyle \text{Answer:}$

$\displaystyle x^2 = 4y$

$\displaystyle \text{Differentiating both sides with respect to } x,$

$\displaystyle 2x = 4\frac{dy}{dx}$

$\displaystyle \Rightarrow \frac{dy}{dx} = \frac{x}{2}$

$\displaystyle \text{Given point } (x_1, y_1) = (2,1)$

$\displaystyle \text{Slope of tangent at } (2,1) = \left.\frac{dy}{dx}\right|_{x=2} = \frac{2}{2} = 1$

$\displaystyle \text{Equation of tangent is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y - 1 = 1(x - 2)$

$\displaystyle \Rightarrow y - 1 = x - 2$

$\displaystyle \Rightarrow x - y - 1 = 0$

$\displaystyle \text{Equation of normal is } y - y_1 = -\frac{1}{m}(x - x_1)$

$\displaystyle \Rightarrow y - 1 = -1(x - 2)$

$\displaystyle \Rightarrow y - 1 = -x + 2$

$\displaystyle \Rightarrow x + y - 3 = 0$

$\displaystyle \text{(xvi) }\;y^2=4x\text{ at }(1,2)$

$\displaystyle \text{Answer:}$

$\displaystyle y^2 = 4x$

$\displaystyle \text{Differentiating both sides with respect to } x,$

$\displaystyle 2y\frac{dy}{dx} = 4$

$\displaystyle \Rightarrow \frac{dy}{dx} = \frac{2}{y}$

$\displaystyle \text{Given point } (x_1, y_1) = (1,2)$

$\displaystyle \text{Slope of tangent at } (1,2) = \left.\frac{dy}{dx}\right|_{y=2} = \frac{2}{2} = 1$

$\displaystyle \text{Equation of tangent is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y - 2 = 1(x - 1)$

$\displaystyle \Rightarrow y - 2 = x - 1$

$\displaystyle \Rightarrow x - y + 1 = 0$

$\displaystyle \text{Equation of normal is } y - y_1 = -\frac{1}{m}(x - x_1)$

$\displaystyle \Rightarrow y - 2 = -1(x - 1)$

$\displaystyle \Rightarrow y - 2 = -x + 1$

$\displaystyle \Rightarrow x + y - 3 = 0$

$\displaystyle \text{(xvii) }\;4x^2+9y^2=36\text{ at }(3\cos\theta,\;2\sin\theta)\;[\text{CBSE 2011}]$

$\displaystyle \text{Answer:}$

$\displaystyle 4x^2 + 9y^2 = 36$

$\displaystyle \text{Differentiating both sides with respect to } x,$

$\displaystyle 8x + 18y\frac{dy}{dx} = 0$

$\displaystyle \Rightarrow 18y\frac{dy}{dx} = -8x$

$\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{4x}{9y}$

$\displaystyle \text{Given point } (x_1, y_1) = (3\cos\theta, 2\sin\theta)$

$\displaystyle \text{Slope of tangent at } (3\cos\theta, 2\sin\theta) = \left.\frac{dy}{dx}\right|_{(3\cos\theta,\,2\sin\theta)} = -\frac{4(3\cos\theta)}{9(2\sin\theta)} = -\frac{2\cos\theta}{3\sin\theta}$

$\displaystyle \text{Equation of tangent is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y - 2\sin\theta = -\frac{2\cos\theta}{3\sin\theta}(x - 3\cos\theta)$

$\displaystyle \Rightarrow 3y\sin\theta - 6\sin^2\theta = -2x\cos\theta + 6\cos^2\theta$

$\displaystyle \Rightarrow 2x\cos\theta + 3y\sin\theta = 6(\cos^2\theta + \sin^2\theta)$

$\displaystyle \Rightarrow 2x\cos\theta + 3y\sin\theta = 6$

$\displaystyle \text{Equation of normal is } y - y_1 = -\frac{1}{m}(x - x_1)$

$\displaystyle \Rightarrow y - 2\sin\theta = \frac{3\sin\theta}{2\cos\theta}(x - 3\cos\theta)$

$\displaystyle \Rightarrow 2y\cos\theta - 4\sin\theta\cos\theta = 3x\sin\theta - 9\sin\theta\cos\theta$

$\displaystyle \Rightarrow 3x\sin\theta - 2y\cos\theta - 5\sin\theta\cos\theta = 0$

$\displaystyle \text{(xviii) }\;y^2=4ax\text{ at }(x_1,y_1)\;[\text{CBSE 2012}]$

$\displaystyle \text{Answer:}$

$\displaystyle y^2 = 4ax$

$\displaystyle \text{Differentiating both sides with respect to } x,$

$\displaystyle 2y\frac{dy}{dx} = 4a$

$\displaystyle \Rightarrow \frac{dy}{dx} = \frac{2a}{y}$

$\displaystyle \text{At the point } (x_1, y_1)$

$\displaystyle \text{Slope of tangent at } (x_1, y_1) = \left.\frac{dy}{dx}\right|_{(x_1,y_1)} = \frac{2a}{y_1} = m$

$\displaystyle \text{Equation of tangent is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y - y_1 = \frac{2a}{y_1}(x - x_1)$

$\displaystyle \Rightarrow y_1y - y_1^2 = 2ax - 2ax_1$

$\displaystyle \Rightarrow y_1y - 4ax_1 = 2ax - 2ax_1$

$\displaystyle \Rightarrow y_1y = 2a(x + x_1)$

$\displaystyle \text{Equation of normal is } y - y_1 = -\frac{1}{m}(x - x_1)$

$\displaystyle \Rightarrow y - y_1 = -\frac{y_1}{2a}(x - x_1)$

$\displaystyle \text{(xix) }\;\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\text{ at }(\sqrt{2}\,a,\;b)\;[\text{CBSE 2014}]$

$\displaystyle \text{Answer:}$

$\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

$\displaystyle \text{Differentiating both sides with respect to } x,$

$\displaystyle \Rightarrow \frac{2x}{a^2} - \frac{2y}{b^2}\frac{dy}{dx} = 0$

$\displaystyle \Rightarrow \frac{2y}{b^2}\frac{dy}{dx} = \frac{2x}{a^2}$

$\displaystyle \Rightarrow \frac{dy}{dx} = \frac{xb^2}{ya^2}$

$\displaystyle \text{Given point } (x_1, y_1) = (\sqrt{2}a, b)$

$\displaystyle \text{Slope of tangent at } (\sqrt{2}a, b) = \left.\frac{dy}{dx}\right|_{(\sqrt{2}a,b)} = \frac{\sqrt{2}a\,b^2}{b\,a^2} = \frac{\sqrt{2}b}{a}$

$\displaystyle \text{Equation of tangent is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y - b = \frac{\sqrt{2}b}{a}(x - \sqrt{2}a)$

$\displaystyle \Rightarrow ay - ab = \sqrt{2}bx - 2ab$

$\displaystyle \Rightarrow \sqrt{2}bx - ay = ab$

$\displaystyle \Rightarrow \frac{\sqrt{2}x}{a} - \frac{y}{b} = 1$

$\displaystyle \text{Equation of normal is } y - y_1 = -\frac{1}{m}(x - x_1)$

$\displaystyle \Rightarrow y - b = -\frac{a}{\sqrt{2}b}(x - \sqrt{2}a)$

$\displaystyle \Rightarrow \sqrt{2}by - \sqrt{2}b^2 = -ax + \sqrt{2}a^2$

$\displaystyle \Rightarrow ax + \sqrt{2}by = \sqrt{2}(a^2 + b^2)$

$\displaystyle \Rightarrow \frac{ax}{\sqrt{2}} + by = a^2 + b^2$

$\displaystyle \textbf{Question 4: }~\text{Find the equation of the tangent to the curve }x=\theta+\sin\theta, \\ y=1+\cos\theta\text{ at }\theta=\frac{\pi}{4}.$
$\displaystyle \text{Answer:}$

$\displaystyle x = \theta + \sin\theta \quad \text{and} \quad y = 1 + \cos\theta$

$\displaystyle \frac{dx}{d\theta} = 1 + \cos\theta \quad \text{and} \quad \frac{dy}{d\theta} = -\sin\theta$

$\displaystyle \therefore \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-\sin\theta}{1 + \cos\theta}$

$\displaystyle \text{Slope of tangent at } \theta = \frac{\pi}{4} = \left.\frac{dy}{dx}\right|_{\theta=\frac{\pi}{4}} = \frac{-\sin\frac{\pi}{4}}{1 + \cos\frac{\pi}{4}} = \frac{-\frac{1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}}}$

$\displaystyle = \frac{-1}{\sqrt{2}+1} = \frac{-1(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)} = 1 - \sqrt{2}$

$\displaystyle \text{Given point } (x_1, y_1) = \left(\frac{\pi}{4} + \sin\frac{\pi}{4},\, 1 + \cos\frac{\pi}{4}\right) = \left(\frac{\pi}{4} + \frac{1}{\sqrt{2}},\, 1 + \frac{1}{\sqrt{2}}\right)$

$\displaystyle \text{Equation of tangent is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y - \left(1 + \frac{1}{\sqrt{2}}\right) = (1 - \sqrt{2}) \left[x - \left(\frac{\pi}{4} + \frac{1}{\sqrt{2}}\right)\right]$

$\displaystyle \textbf{Question 5: }~\text{Find the equations of the tangent and the normal to the following curves } \\ \text{at the indicated points:}$
$\displaystyle \text{(i) }\;x=\theta+\sin\theta,\;y=1+\cos\theta\text{ at }\theta=\frac{\pi}{2}.$
$\displaystyle \text{Answer:}$

$\displaystyle x = \theta + \sin\theta \quad \text{and} \quad y = 1 + \cos\theta$

$\displaystyle \frac{dx}{d\theta} = 1 + \cos\theta \quad \text{and} \quad \frac{dy}{d\theta} = -\sin\theta$

$\displaystyle \therefore \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-\sin\theta}{1 + \cos\theta}$

$\displaystyle \text{Slope of tangent at } \theta = \frac{\pi}{2} = \left.\frac{dy}{dx}\right|_{\theta=\frac{\pi}{2}} = \frac{-\sin\frac{\pi}{2}}{1 + \cos\frac{\pi}{2}} = \frac{-1}{1 + 0} = -1$

$\displaystyle \text{Given point } (x_1, y_1) = \left(\frac{\pi}{2} + \sin\frac{\pi}{2},\, 1 + \cos\frac{\pi}{2}\right) = \left(\frac{\pi}{2} + 1,\, 1\right)$

$\displaystyle \text{Equation of tangent is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y - 1 = -1\left(x - \frac{\pi}{2} - 1\right)$

$\displaystyle \Rightarrow y - 1 = -x + \frac{\pi}{2} + 1$

$\displaystyle \Rightarrow x + y - \frac{\pi}{2} - 2 = 0$

$\displaystyle \text{Equation of normal is } y - y_1 = -\frac{1}{m}(x - x_1)$

$\displaystyle \Rightarrow y - 1 = 1\left(x - \frac{\pi}{2} - 1\right)$

$\displaystyle \Rightarrow y - 1 = x - \frac{\pi}{2} - 1$

$\displaystyle \Rightarrow 2y - 2 = 2x - \pi - 2$

$\displaystyle \Rightarrow 2x - 2y - \pi = 0$

$\displaystyle \text{(ii) }\;x=\frac{2at^2}{1+t^2},\;y=\frac{2at^3}{1+t^2}\text{ at }t=\frac{1}{2}.$
$\displaystyle \text{Answer:}$

$\displaystyle x = \frac{2at^2}{1+t^2} \quad \text{and} \quad y = \frac{2at^3}{1+t^2}$

$\displaystyle \frac{dx}{dt} = \frac{(1+t^2)(4at) - 2at^2(2t)}{(1+t^2)^2} = \frac{4at}{(1+t^2)^2}$

$\displaystyle \frac{dy}{dt} = \frac{(1+t^2)(6at^2) - 2at^3(2t)}{(1+t^2)^2} = \frac{6at^2 + 2at^4}{(1+t^2)^2}$

$\displaystyle \therefore \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{6at^2 + 2at^4}{4at} = \frac{a(3 + t^2)}{2t}$

$\displaystyle \text{Slope of tangent at } t = \frac{1}{2} = \left.\frac{dy}{dx}\right|_{t=\frac{1}{2}} = \frac{a\left(3 + \frac{1}{4}\right)}{2 \cdot \frac{1}{2}} = \frac{13a}{8} \cdot \frac{1}{2a} = \frac{13}{16}$

$\displaystyle \text{Given point } (x_1, y_1) = \left(\frac{2a\left(\frac{1}{2}\right)^2}{1+\left(\frac{1}{2}\right)^2}, \frac{2a\left(\frac{1}{2}\right)^3}{1+\left(\frac{1}{2}\right)^2}\right) = \left(\frac{2a}{5}, \frac{a}{5}\right)$

$\displaystyle \text{Equation of tangent is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y - \frac{a}{5} = \frac{13}{16}\left(x - \frac{2a}{5}\right)$

$\displaystyle \Rightarrow 16y - \frac{16a}{5} = 13x - \frac{26a}{5}$

$\displaystyle \Rightarrow 65x - 80y - 10a = 0$

$\displaystyle \Rightarrow 13x - 16y - 2a = 0$

$\displaystyle \text{Equation of normal is } y - y_1 = -\frac{1}{m}(x - x_1)$

$\displaystyle \Rightarrow y - \frac{a}{5} = -\frac{16}{13}\left(x - \frac{2a}{5}\right)$

$\displaystyle \Rightarrow 13y - \frac{13a}{5} = -16x + \frac{32a}{5}$

$\displaystyle \Rightarrow 16x + 13y - 9a = 0$

$\displaystyle \text{(iii) }\;x=at^2,\;y=2at\text{ at }t=1.$
$\displaystyle \text{Answer:}$

$\displaystyle x = at^2 \quad \text{and} \quad y = 2at$

$\displaystyle \frac{dx}{dt} = 2at \quad \text{and} \quad \frac{dy}{dt} = 2a$

$\displaystyle \therefore \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2a}{2at} = \frac{1}{t}$

$\displaystyle \text{Slope of tangent at } t = 1 = \left.\frac{dy}{dx}\right|_{t=1} = \frac{1}{1} = 1$

$\displaystyle \text{Now, } (x_1, y_1) = (a, 2a)$

$\displaystyle \text{Equation of tangent is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y - 2a = 1(x - a)$

$\displaystyle \Rightarrow y - 2a = x - a$

$\displaystyle \Rightarrow x - y + a = 0$

$\displaystyle \text{Equation of normal is } y - y_1 = -\frac{1}{m}(x - x_1)$

$\displaystyle \Rightarrow y - 2a = -1(x - a)$

$\displaystyle \Rightarrow y - 2a = -x + a$

$\displaystyle \Rightarrow x + y - 3a = 0$

$\displaystyle \text{(iv) }\;x=a\sec t,\;y=b\tan t\text{ at }t.$
$\displaystyle \text{Answer:}$

$\displaystyle x = a\sec t \quad \text{and} \quad y = b\tan t$

$\displaystyle \frac{dx}{dt} = a\sec t\tan t \quad \text{and} \quad \frac{dy}{dt} = b\sec^2 t$

$\displaystyle \therefore \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{b\sec^2 t}{a\sec t\tan t} = \frac{b}{a}\,\mathrm{cosec}\,t$

$\displaystyle \text{Slope of tangent at parameter } t = \left.\frac{dy}{dx}\right|_{t} = \frac{b}{a}\,\mathrm{cosec}\,t$

$\displaystyle \text{Given point } (x_1, y_1) = (a\sec t,\, b\tan t)$

$\displaystyle \text{Equation of tangent is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y - b\tan t = \frac{b}{a}\,\mathrm{cosec}\,t\,(x - a\sec t)$

$\displaystyle \Rightarrow y - \frac{b\sin t}{\cos t} = \frac{b}{a\sin t}\left(x - \frac{a}{\cos t}\right)$

$\displaystyle \Rightarrow y\cos t - b\sin t = \frac{b}{a\sin t}(x\cos t - a)$

$\displaystyle \Rightarrow ay\sin t\cos t - ab\sin^2 t = bx\cos t - ab$

$\displaystyle \Rightarrow bx\cos t - ay\sin t\cos t = ab(1 - \sin^2 t)$

$\displaystyle \Rightarrow bx\cos t - ay\sin t\cos t = ab\cos^2 t$

$\displaystyle \Rightarrow bx\sec t - ay\tan t = ab$

$\displaystyle \text{Equation of normal is } y - y_1 = -\frac{1}{m}(x - x_1)$

$\displaystyle \Rightarrow y - b\tan t = -\frac{a}{b}\sin t\,(x - a\sec t)$

$\displaystyle \Rightarrow y\cos t - b\sin t = -\frac{a}{b}\sin t\,(x\cos t - a)$

$\displaystyle \Rightarrow by\cos t - b^2\sin t = -ax\sin t\cos t + a^2\sin t$

$\displaystyle \Rightarrow ax\sin t\cos t + by\cos t = (a^2 + b^2)\sin t$

$\displaystyle \Rightarrow ax\cos t + by\,\mathrm{cot}\,t = a^2 + b^2$

$\displaystyle \text{(v) }\;x=a(\theta+\sin\theta),\;y=a(1-\cos\theta)\text{ at }\theta.$
$\displaystyle \text{Answer:}$

$\displaystyle x = a(\theta + \sin\theta) \quad \text{and} \quad y = a(1 - \cos\theta)$

$\displaystyle \frac{dx}{d\theta} = a(1 + \cos\theta) \quad \text{and} \quad \frac{dy}{d\theta} = a\sin\theta$

$\displaystyle \therefore \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{a\sin\theta}{a(1 + \cos\theta)} = \frac{\sin\theta}{1 + \cos\theta}$

$\displaystyle = \frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}} = \tan\frac{\theta}{2} \quad (1)$

$\displaystyle \text{Slope of tangent at parameter } \theta = \left.\frac{dy}{dx}\right|_{\theta} = \tan\frac{\theta}{2}$

$\displaystyle \text{Now, } (x_1, y_1) = \bigl(a(\theta + \sin\theta),\, a(1 - \cos\theta)\bigr)$

$\displaystyle \text{Equation of tangent is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y - a(1 - \cos\theta) = \tan\frac{\theta}{2}\left[x - a(\theta + \sin\theta)\right]$

$\displaystyle \Rightarrow y - a\left(2\sin^2\frac{\theta}{2}\right) = x\tan\frac{\theta}{2} - a\theta\tan\frac{\theta}{2} - a\sin\theta\tan\frac{\theta}{2}$

$\displaystyle \Rightarrow y - 2a\sin^2\frac{\theta}{2} = (x - a\theta)\tan\frac{\theta}{2} - 2a\sin^2\frac{\theta}{2}$

$\displaystyle \Rightarrow y = (x - a\theta)\tan\frac{\theta}{2}$

$\displaystyle \text{Equation of normal is } y - y_1 = -\frac{1}{m}(x - x_1)$

$\displaystyle \Rightarrow y - a(1 - \cos\theta) = -\cot\frac{\theta}{2}\left[x - a(\theta + \sin\theta)\right]$

$\displaystyle \Rightarrow \tan\frac{\theta}{2} \left[y - a\left(2\sin^2\frac{\theta}{2}\right)\right] = -x + a\theta + a\sin\theta$

$\displaystyle \Rightarrow \tan\frac{\theta}{2}(y - 2a) + a\sin\theta = -x + a\theta + a\sin\theta$

$\displaystyle \Rightarrow \tan\frac{\theta}{2}(y - 2a) = -x + a\theta$

$\displaystyle \Rightarrow x + \tan\frac{\theta}{2}(y - 2a) - a\theta = 0$

$\displaystyle \text{(vi) }\;x=3\cos\theta-\cos^3\theta,\;y=3\sin\theta-\sin^3\theta.\;[\text{CBSE 2016}]$
$\displaystyle \text{Answer:}$

$\displaystyle x = 3\cos\theta - \cos^3\theta \quad \text{and} \quad y = 3\sin\theta - \sin^3\theta$

$\displaystyle \frac{dx}{d\theta} = -3\sin\theta + 3\cos^2\theta\sin\theta$

$\displaystyle \frac{dy}{d\theta} = 3\cos\theta - 3\sin^2\theta\cos\theta$

$\displaystyle \therefore \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{3\cos\theta(1-\sin^2\theta)}{-3\sin\theta(1-\cos^2\theta)} = \frac{\cos^3\theta}{-\sin^3\theta} = -\tan^3\theta$

$\displaystyle \text{Slope of tangent at parameter } \theta = -\tan^3\theta$

$\displaystyle \text{Equation of tangent at } \theta \text{ is }$

$\displaystyle y - (3\sin\theta - \sin^3\theta) = -\tan^3\theta\,[\,x - (3\cos\theta - \cos^3\theta)\,]$

$\displaystyle \Rightarrow 4\bigl(y\cos^3\theta - x\sin^3\theta\bigr) = 3\sin4\theta$

$\displaystyle \text{Equation of normal at } \theta \text{ is }$

$\displaystyle y - (3\sin\theta - \sin^3\theta) = \frac{1}{\tan^3\theta}\,[\,x - (3\cos\theta - \cos^3\theta)\,]$

$\displaystyle \Rightarrow y\sin^3\theta - x\cos^3\theta = 3\sin^4\theta - \sin^6\theta - 3\cos^4\theta + \cos^6\theta$

$\displaystyle \textbf{Question 6: }~\text{Find the equation of the normal to the curve } \\ x^2+2y^2-4x-6y+8=0\text{ at the point whose abscissa is }2.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Abscissa means the horizontal coordinate of a point.}$

$\displaystyle \text{Given that abscissa } = 2 \Rightarrow x = 2$

$\displaystyle x^2 + 2y^2 - 4x - 6y + 8 = 0 \quad (1)$

$\displaystyle \text{Differentiating both sides with respect to } x,$

$\displaystyle 2x + 4y\frac{dy}{dx} - 4 - 6\frac{dy}{dx} = 0$

$\displaystyle \Rightarrow \frac{dy}{dx}(4y - 6) = 4 - 2x$

$\displaystyle \Rightarrow \frac{dy}{dx} = \frac{4 - 2x}{4y - 6} = \frac{2 - x}{2y - 3}$

$\displaystyle \text{When } x = 2,\ \text{from (1) we get}$

$\displaystyle 4 + 2y^2 - 8 - 6y + 8 = 0$

$\displaystyle \Rightarrow 2y^2 - 6y + 4 = 0$

$\displaystyle \Rightarrow y^2 - 3y + 2 = 0$

$\displaystyle \Rightarrow (y - 1)(y - 2) = 0$

$\displaystyle \Rightarrow y = 1 \ \text{or} \ y = 2$

$\displaystyle \text{Case 1: } y = 1$

$\displaystyle \text{Slope of tangent at } (2,1) = \left.\frac{dy}{dx}\right|_{(2,1)} = \frac{2 - 2}{2(1) - 3} = 0$

$\displaystyle (x_1, y_1) = (2,1)$

$\displaystyle \text{Slope of normal is undefined since } m = 0$

$\displaystyle \text{Equation of normal is }$

$\displaystyle \text{Since slope of tangent } m = 0,$

$\displaystyle \text{the normal is vertical.}$

$\displaystyle \Rightarrow \text{Equation of normal is } x = 2$

$\displaystyle \text{Case 2: } y = 2$

$\displaystyle \text{Slope of tangent at } (2,2) = \left.\frac{dy}{dx}\right|_{(2,2)} = \frac{0}{1} = 0$

$\displaystyle (x_1, y_1) = (2,2)$

$\displaystyle \text{Since slope of tangent } m = 0,$

$\displaystyle \text{the normal is vertical.}$

$\displaystyle \Rightarrow \text{Equation of normal is } x = 2$

$\displaystyle \text{In both cases, the equation of the normal is } x = 2$

$\displaystyle \textbf{Question 7: }~\text{Find the equation of the normal to the curve }ay^2=x^3\text{ at the point } \\ (am^2,\;am^3).\hspace{5.0cm} \;[\text{CBSE 2012}]$
$\displaystyle \text{Answer:}$

$\displaystyle ay^2 = x^3$

$\displaystyle \text{Differentiating both sides with respect to } x,$

$\displaystyle 2ay\frac{dy}{dx} = 3x^2$

$\displaystyle \Rightarrow \frac{dy}{dx} = \frac{3x^2}{2ay}$

$\displaystyle \text{Given point } (x_1, y_1) = (am^2, am^3)$

$\displaystyle \text{Slope of tangent at } (am^2, am^3) = \left.\frac{dy}{dx}\right|_{(am^2,am^3)} = \frac{3a^2m^4}{2a^2m^3} = \frac{3m}{2}$

$\displaystyle \text{Slope of normal} = -\frac{2}{3m}$

$\displaystyle \text{Equation of normal is } y - y_1 = -\frac{1}{m_t}(x - x_1)$

$\displaystyle \Rightarrow y - am^3 = -\frac{2}{3m}(x - am^2)$

$\displaystyle \Rightarrow 3my - 3am^4 = -2x + 2am^2$

$\displaystyle \Rightarrow 2x + 3my - am^2(2 + 3m^2) = 0$

$\displaystyle \textbf{Question 8: }~\text{The equation of the tangent at }(2,3)\text{ on the curve } \\ y^2=ax^3+b\text{ is }y=4x-5.\text{ Find the values of }a\text{ and }b.\hspace{5.0cm} \;[\text{CBSE 2016}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The slope of the given line } y = 4x - 5 \text{ is } 4$

$\displaystyle y^2 = ax^3 + b \quad (1)$

$\displaystyle \text{Differentiating both sides with respect to } x,$

$\displaystyle 2y\frac{dy}{dx} = 3ax^2$

$\displaystyle \Rightarrow \frac{dy}{dx} = \frac{3ax^2}{2y}$

$\displaystyle \text{Slope of tangent at } (2,3) = \left.\frac{dy}{dx}\right|_{(2,3)} = \frac{3a(2)^2}{2(3)} = \frac{12a}{6} = 2a$

$\displaystyle \text{Given that slope of tangent equals slope of given line,}$

$\displaystyle 2a = 4$

$\displaystyle \Rightarrow a = 2$

$\displaystyle \text{Substituting } a = 2,\ x = 2,\ y = 3 \text{ in (1),}$

$\displaystyle 3^2 = 2(2)^3 + b$

$\displaystyle \Rightarrow 9 = 16 + b$

$\displaystyle \Rightarrow b = -7$

$\displaystyle \text{Hence, } a = 2 \text{ and } b = -7$

$\displaystyle \textbf{Question 9: }~\text{Find the equation of the tangent line to the curve } \\ y=x^2+4x-16\text{ which is parallel to the line }3x-y+1=0.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } (x_0, y_0) \text{ be the point of contact of the tangent with the curve.}$

$\displaystyle y = x^2 + 4x - 16$

$\displaystyle \text{Since } (x_0, y_0) \text{ lies on the curve,}$

$\displaystyle y_0 = x_0^2 + 4x_0 - 16 \quad (1)$

$\displaystyle \text{Differentiating } y = x^2 + 4x - 16 \text{ with respect to } x,$

$\displaystyle \frac{dy}{dx} = 2x + 4$

$\displaystyle \text{Slope of tangent at } (x_0, y_0) = \left.\frac{dy}{dx}\right|_{(x_0,y_0)} = 2x_0 + 4$

$\displaystyle \text{Given that the tangent is parallel to the line of slope } 3,$

$\displaystyle \text{Slope of tangent} = \text{slope of given line}$

$\displaystyle 2x_0 + 4 = 3$

$\displaystyle \Rightarrow 2x_0 = -1$

$\displaystyle \Rightarrow x_0 = -\frac{1}{2}$

$\displaystyle \text{From (1),}$

$\displaystyle y_0 = \left(-\frac{1}{2}\right)^2 + 4\left(-\frac{1}{2}\right) - 16$

$\displaystyle \Rightarrow y_0 = \frac{1}{4} - 2 - 16 = -\frac{71}{4}$

$\displaystyle \text{Thus, } (x_0, y_0) = \left(-\frac{1}{2}, -\frac{71}{4}\right)$

$\displaystyle \text{Equation of tangent is } y - y_0 = m(x - x_0)$

$\displaystyle \Rightarrow y + \frac{71}{4} = 3\left(x + \frac{1}{2}\right)$

$\displaystyle \Rightarrow \frac{4y + 71}{4} = \frac{3(2x + 1)}{2}$

$\displaystyle \Rightarrow 4y + 71 = 12x + 6$

$\displaystyle \Rightarrow 12x - 4y - 65 = 0$

$\displaystyle \textbf{Question 10: }~\text{Find an equation of normal line to the curve } \\ y=x^3+2x+6\text{ which is parallel to the line }x+14y+4=0.\hspace{1.0cm} \;[\text{CBSE 2013}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } (x_1, y_1) \text{ be a point on the curve at which the normal is to be found.}$

$\displaystyle \text{Slope of the given line } = -\frac{1}{14}$

$\displaystyle \text{Given curve: } y = x^3 + 2x + 6$

$\displaystyle \text{Since } (x_1, y_1) \text{ lies on the curve,}$

$\displaystyle y_1 = x_1^3 + 2x_1 + 6$

$\displaystyle \text{Differentiating } y = x^3 + 2x + 6 \text{ with respect to } x,$

$\displaystyle \frac{dy}{dx} = 3x^2 + 2$

$\displaystyle \text{Slope of tangent at } (x_1, y_1) = \left.\frac{dy}{dx}\right|_{(x_1,y_1)} = 3x_1^2 + 2$

$\displaystyle \text{Slope of normal} = -\frac{1}{3x_1^2 + 2}$

$\displaystyle \text{Given that slope of normal equals slope of the given line,}$

$\displaystyle -\frac{1}{3x_1^2 + 2} = -\frac{1}{14}$

$\displaystyle \Rightarrow 3x_1^2 + 2 = 14$

$\displaystyle \Rightarrow 3x_1^2 = 12$

$\displaystyle \Rightarrow x_1^2 = 4$

$\displaystyle \Rightarrow x_1 = \pm 2$

$\displaystyle \text{Case 1: } x_1 = 2$

$\displaystyle y_1 = 2^3 + 2(2) + 6 = 18$

$\displaystyle (x_1, y_1) = (2,18)$

$\displaystyle \text{Equation of normal is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y - 18 = -\frac{1}{14}(x - 2)$

$\displaystyle \Rightarrow 14y - 252 = -x + 2$

$\displaystyle \Rightarrow x + 14y - 254 = 0$

$\displaystyle \text{Case 2: } x_1 = -2$

$\displaystyle y_1 = (-2)^3 + 2(-2) + 6 = -6$

$\displaystyle (x_1, y_1) = (-2,-6)$

$\displaystyle \text{Equation of normal is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y + 6 = -\frac{1}{14}(x + 2)$

$\displaystyle \Rightarrow 14y + 84 = -x - 2$

$\displaystyle \Rightarrow x + 14y + 86 = 0$

$\displaystyle \textbf{Question 11: }~\text{Determine the equation(s) of tangent(s) line to the curve } \\ y=4x^3-3x+5\text{ which are perpendicular to the line }9y+x+3=0.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } (x_1, y_1) \text{ be a point on the curve at which the tangent is to be drawn.}$

$\displaystyle \text{Slope of the given line } = -\frac{1}{9}$

$\displaystyle \text{Since the tangent is perpendicular to the given line,}$

$\displaystyle \text{Slope of tangent} = -\frac{1}{\left(-\frac{1}{9}\right)} = 9$

$\displaystyle \text{Given curve: } y = 4x^3 - 3x + 5$

$\displaystyle \text{Since } (x_1, y_1) \text{ lies on the curve,}$

$\displaystyle y_1 = 4x_1^3 - 3x_1 + 5$

$\displaystyle \text{Differentiating } y = 4x^3 - 3x + 5 \text{ with respect to } x,$

$\displaystyle \frac{dy}{dx} = 12x^2 - 3$

$\displaystyle \text{Slope of tangent at } (x_1, y_1) = \left.\frac{dy}{dx}\right|_{(x_1,y_1)} = 12x_1^2 - 3$

$\displaystyle \text{Given that slope of tangent equals } 9,$

$\displaystyle 12x_1^2 - 3 = 9$

$\displaystyle \Rightarrow 12x_1^2 = 12$

$\displaystyle \Rightarrow x_1^2 = 1$

$\displaystyle \Rightarrow x_1 = \pm 1$

$\displaystyle \text{Case 1: } x_1 = 1$

$\displaystyle y_1 = 4(1)^3 - 3(1) + 5 = 6$

$\displaystyle (x_1, y_1) = (1,6)$

$\displaystyle \text{Equation of tangent is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y - 6 = 9(x - 1)$

$\displaystyle \Rightarrow y - 6 = 9x - 9$

$\displaystyle \Rightarrow 9x - y - 3 = 0$

$\displaystyle \text{Case 2: } x_1 = -1$

$\displaystyle y_1 = 4(-1)^3 - 3(-1) + 5 = 4$

$\displaystyle (x_1, y_1) = (-1,4)$

$\displaystyle \text{Equation of tangent is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y - 4 = 9(x + 1)$

$\displaystyle \Rightarrow y - 4 = 9x + 9$

$\displaystyle \Rightarrow 9x - y + 13 = 0$

$\displaystyle \textbf{Question 12: }~\text{Find the equation of a normal to the curve }y=x\log_e x\text{ which is } \\ \text{parallel to the line }2x-2y+3=0.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Slope of the given line is } 1$

$\displaystyle \text{Let } (x_1, y_1) \text{ be the point where the normal is drawn to the curve.}$

$\displaystyle \text{Given curve: } y = x\log_e x$

$\displaystyle \text{Since } (x_1, y_1) \text{ lies on the curve,}$

$\displaystyle y_1 = x_1 \log_e x_1 \quad (1)$

$\displaystyle \text{Differentiating } y = x\log_e x \text{ with respect to } x,$

$\displaystyle \frac{dy}{dx} = x\cdot\frac{1}{x} + \log_e x \cdot 1 = 1 + \log_e x$

$\displaystyle \text{Slope of tangent at } (x_1, y_1) = 1 + \log_e x_1$

$\displaystyle \text{Slope of normal} = -\frac{1}{1 + \log_e x_1}$

$\displaystyle \text{Given that slope of normal equals slope of the given line,}$

$\displaystyle -\frac{1}{1 + \log_e x_1} = 1$

$\displaystyle \Rightarrow -1 = 1 + \log_e x_1$

$\displaystyle \Rightarrow \log_e x_1 = -2$

$\displaystyle \Rightarrow x_1 = e^{-2} = \frac{1}{e^2}$

$\displaystyle \text{From (1),}$

$\displaystyle y_1 = \frac{1}{e^2}\log_e\!\left(\frac{1}{e^2}\right) = \frac{1}{e^2}(-2) = -\frac{2}{e^2}$

$\displaystyle \text{Thus, } (x_1, y_1) = \left(\frac{1}{e^2}, -\frac{2}{e^2}\right)$

$\displaystyle \text{Equation of normal is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y + \frac{2}{e^2} = 1\left(x - \frac{1}{e^2}\right)$

$\displaystyle \Rightarrow y + \frac{2}{e^2} = x - \frac{1}{e^2}$

$\displaystyle \Rightarrow x - y = \frac{3}{e^2}$

$\displaystyle \Rightarrow x - y = 3e^{-2}$

$\displaystyle \textbf{Question 13: }~\text{Find the equation of the tangent line to the curve } \\ y=x^2-2x+7\text{ which is }(i)\text{ parallel to the line }2x-y+9=0\;(ii)\text{ perpendicular to the line } \\ 5y-15x-13=0.\hspace{5.0cm} \;[\text{CBSE 2014}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }$

$\displaystyle \text{The equation of the given curve is } y = x^2 - 2x + 7$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = 2x - 2$

$\displaystyle \text{The equation of the given line is } 2x - y + 9 = 0$

$\displaystyle \Rightarrow y = 2x + 9$

$\displaystyle \text{This is of the form } y = mx + c$

$\displaystyle \therefore \text{slope of the given line } = 2$

$\displaystyle \text{If a tangent is parallel to the given line,}$

$\displaystyle \text{slope of tangent} = \text{slope of given line}$

$\displaystyle \Rightarrow 2x - 2 = 2$

$\displaystyle \Rightarrow 2x = 4$

$\displaystyle \Rightarrow x = 2$

$\displaystyle \text{Now, when } x = 2,$

$\displaystyle y = 2^2 - 2(2) + 7 = 7$

$\displaystyle \text{Thus, the point of contact is } (2,7)$

$\displaystyle \text{Equation of the tangent is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y - 7 = 2(x - 2)$

$\displaystyle \Rightarrow y - 7 = 2x - 4$

$\displaystyle \Rightarrow y - 2x - 3 = 0$

$\displaystyle \text{Hence, the equation of the tangent is } y - 2x - 3 = 0$

$\displaystyle \text{(ii) }$

$\displaystyle \text{The equation of the given curve is } y = x^2 - 2x + 7$

$\displaystyle \text{Differentiating with respect to } x,$

$\displaystyle \frac{dy}{dx} = 2x - 2$

$\displaystyle \text{The equation of the given line is } 5y - 15x = 13$

$\displaystyle \Rightarrow 5y = 15x + 13$

$\displaystyle \Rightarrow y = 3x + \frac{13}{5}$

$\displaystyle \text{This is of the form } y = mx + c$

$\displaystyle \therefore \text{slope of the given line } = 3$

$\displaystyle \text{If a tangent is perpendicular to the given line,}$

$\displaystyle \text{slope of tangent} = -\frac{1}{\text{slope of given line}} = -\frac{1}{3}$

$\displaystyle \text{At the point of contact,}$

$\displaystyle 2x - 2 = -\frac{1}{3}$

$\displaystyle \Rightarrow 2x = 2 - \frac{1}{3} = \frac{5}{3}$

$\displaystyle \Rightarrow x = \frac{5}{6}$

$\displaystyle \text{Now, when } x = \frac{5}{6},$

$\displaystyle y = \left(\frac{5}{6}\right)^2 - 2\left(\frac{5}{6}\right) + 7$

$\displaystyle \Rightarrow y = \frac{25}{36} - \frac{10}{6} + 7$

$\displaystyle \Rightarrow y = \frac{25 - 60 + 252}{36} = \frac{217}{36}$

$\displaystyle \text{Thus, the point of contact is } \left(\frac{5}{6}, \frac{217}{36}\right)$

$\displaystyle \text{Equation of the tangent is } y - y_1 = m(x - x_1)$

$\displaystyle \Rightarrow y - \frac{217}{36} = -\frac{1}{3}\left(x - \frac{5}{6}\right)$

$\displaystyle \Rightarrow 36y - 217 = -12x + 10$

$\displaystyle \Rightarrow 36y + 12x - 227 = 0$

$\displaystyle \text{Hence, the equation of the tangent is } 36y + 12x - 227 = 0$

$\displaystyle \textbf{Question 14: }~\text{Find the equations of all lines having slope }2\text{ and that are } \\ \text{tangent to the curve }y=\frac{1}{x-3},\;x\ne 3.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } (x_1, y_1) \text{ be the point where the tangent is drawn to the curve.}$

$\displaystyle \text{Given curve: } y = \frac{1}{x - 3}$

$\displaystyle \text{Since } (x_1, y_1) \text{ lies on the curve,}$

$\displaystyle y_1 = \frac{1}{x_1 - 3}$

$\displaystyle \text{Differentiating } y = \frac{1}{x - 3} \text{ with respect to } x,$

$\displaystyle \frac{dy}{dx} = -\frac{1}{(x - 3)^2}$

$\displaystyle \text{Slope of tangent at } (x_1, y_1) = \left.\frac{dy}{dx}\right|_{x = x_1} = -\frac{1}{(x_1 - 3)^2}$

$\displaystyle \text{Given that slope of the tangent is } 2$

$\displaystyle -\frac{1}{(x_1 - 3)^2} = 2$

$\displaystyle \Rightarrow (x_1 - 3)^2 = -\frac{1}{2}$

$\displaystyle \text{This is not possible since } (x_1 - 3)^2 \ge 0$

$\displaystyle \text{Hence, there does not exist any tangent to the curve having slope } 2$

$\displaystyle \textbf{Question 15: }~\text{Find the equations of all lines of slope zero and that are } \\ \text{tangent to the curve }y=\frac{1}{x^2-2x+3}.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Slope of the given tangent is } 0.$

$\displaystyle \text{Let } (x_1,y_1) \text{ be the point where the tangent is drawn to the curve.}$

$\displaystyle \text{Since the point lies on the curve,}$

$\displaystyle y_1=\frac{1}{x_1^2-2x_1+3}\; \text{......(1)}$

$\displaystyle \text{Now, } y=\frac{1}{x^2-2x+3}$

$\displaystyle \frac{dy}{dx} =\frac{(x^2-2x+3)(0)-(2x-2)(1)}{(x^2-2x+3)^2} =\frac{-2x+2}{(x^2-2x+3)^2}$

$\displaystyle \text{Slope of tangent } =\left(\frac{dy}{dx}\right)_{(x_1,y_1)} =\frac{-2x_1+2}{(x_1^2-2x_1+3)^2}$

$\displaystyle \text{Given that slope of tangent }=0,$

$\displaystyle \frac{-2x_1+2}{(x_1^2-2x_1+3)^2}=0$

$\displaystyle -2x_1+2=0$

$\displaystyle 2x_1=2$

$\displaystyle x_1=1$

$\displaystyle \text{Now, from (1),}$

$\displaystyle y_1=\frac{1}{1-2+3}=\frac{1}{2}$

$\displaystyle \therefore (x_1,y_1)=\left(1,\frac{1}{2}\right)$

$\displaystyle \text{Equation of the tangent is,}$

$\displaystyle y-y_1=m(x-x_1)$

$\displaystyle y-\frac{1}{2}=0(x-1)$

$\displaystyle y=\frac{1}{2}$

$\displaystyle \textbf{Question 16: }~\text{Find the equation of the tangent to the curve }y=\sqrt{3x-2}\text{ which } \\ \text{is parallel to the line }4x-2y+5=0.\hspace{5.0cm} \;[\text{CBSE 2005, 2009}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Slope of the given line is }2.$

$\displaystyle \text{Let }(x_1,y_1)\text{ be the point where the tangent is drawn to the curve }y=\sqrt{3x-2}.$

$\displaystyle \text{Since the point lies on the curve,}$

$\displaystyle y_1=\sqrt{3x_1-2}\quad\text{...(1)}$

$\displaystyle \text{Now, }y=\sqrt{3x-2}.$

$\displaystyle \frac{dy}{dx}=\frac{3}{2\sqrt{3x-2}}.$

$\displaystyle \text{Slope of tangent at }(x_1,y_1)=\frac{3}{2\sqrt{3x_1-2}}.$

$\displaystyle \text{Given that slope of tangent = slope of the given line,}$

$\displaystyle \frac{3}{2\sqrt{3x_1-2}}=2.$

$\displaystyle 3=4\sqrt{3x_1-2}.$

$\displaystyle 9=16(3x_1-2).$

$\displaystyle \frac{9}{16}=3x_1-2.$

$\displaystyle 3x_1=\frac{9}{16}+2=\frac{41}{16}.$

$\displaystyle x_1=\frac{41}{48}.$

$\displaystyle \text{Now, }y_1=\sqrt{3\left(\frac{41}{48}\right)-2} =\sqrt{\frac{123}{48}-\frac{96}{48}} =\sqrt{\frac{27}{48}} =\sqrt{\frac{9}{16}} =\frac{3}{4}\quad\text{[From (1)]}.$

$\displaystyle \therefore (x_1,y_1)=\left(\frac{41}{48},\frac{3}{4}\right).$

$\displaystyle \text{Equation of the tangent is,}$

$\displaystyle y-y_1=m(x-x_1).$

$\displaystyle y-\frac{3}{4}=2\left(x-\frac{41}{48}\right).$

$\displaystyle \frac{4y-3}{4}=2\left(\frac{48x-41}{48}\right).$

$\displaystyle 24y-18=48x-41.$

$\displaystyle 48x-24y-23=0.$

$\displaystyle \textbf{Question 17: }~\text{Find the equation of the tangent to the curve } \\ x^2+3y-3=0,\text{ which is parallel to the line }y=4x-5.\hspace{5.0cm} \;[\text{CBSE 2005}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Suppose }(x_1,y_1)\text{ be the point of contact of the tangent.}$

$\displaystyle \text{We can find the slope of the given line by differentiating its equation w.r.t. }x.$

$\displaystyle \text{So, slope of the line }=4.$

$\displaystyle \text{Since }(x_1,y_1)\text{ lies on the curve, therefore}$

$\displaystyle x_1^2+3y_1-3=0\quad\text{...(1)}$

$\displaystyle \text{Now, }x^2+3y-3=0.$

$\displaystyle 2x+3\frac{dy}{dx}=0.$

$\displaystyle \frac{dy}{dx}=-\frac{2x}{3}.$

$\displaystyle \text{Slope of tangent, }m=\left(\frac{dy}{dx}\right)_{(x_1,y_1)}=-\frac{2x_1}{3}.$

$\displaystyle \text{Given that the tangent is parallel to the line,}$

$\displaystyle \text{slope of tangent = slope of the given line.}$

$\displaystyle -\frac{2x_1}{3}=4.$

$\displaystyle x_1=-6.$

$\displaystyle \text{From (1),}$

$\displaystyle 36+3y_1-3=0.$

$\displaystyle 3y_1=-33.$

$\displaystyle y_1=-11.$

$\displaystyle \therefore (x_1,y_1)=(-6,-11).$

$\displaystyle \text{Equation of tangent is,}$

$\displaystyle y-y_1=m(x-x_1).$

$\displaystyle y+11=4(x+6).$

$\displaystyle y+11=4x+24.$

$\displaystyle 4x-y+13=0.$

$\displaystyle \textbf{Question 18: }~\text{Prove that }\left(\frac{x}{a}\right)^n+\left(\frac{y}{b}\right)^n=2\text{ touches the straight line } \\ \frac{x}{a}+\frac{y}{b}=2\text{ for all }n\in\mathbb{N},\text{ at the point }(a,b).$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Now, }\left(\frac{x}{a}\right)^n+\left(\frac{y}{b}\right)^n=2.$

$\displaystyle \text{Differentiating both sides w.r.t. }x,$

$\displaystyle \frac{n}{a}\left(\frac{x}{a}\right)^{n-1} +\frac{n}{b}\left(\frac{y}{b}\right)^{\,n-1}\frac{dy}{dx}=0.$

$\displaystyle \frac{n}{b}\left(\frac{y}{b}\right)^{\,n-1}\frac{dy}{dx} =-\frac{n}{a}\left(\frac{x}{a}\right)^{n-1}.$

$\displaystyle \frac{dy}{dx} =-\frac{n}{a}\left(\frac{x}{a}\right)^{n-1} \times \frac{b}{n}\left(\frac{b}{y}\right)^{n-1} =-\frac{b}{a}\left(\frac{bx}{ay}\right)^{n-1}.$

$\displaystyle \text{Slope of tangent }= \left(\frac{dy}{dx}\right)_{(a,b)} =-\frac{b}{a}\left(\frac{b\times a}{a\times b}\right)^{n-1} =-\frac{b}{a}\quad\text{...(2)}.$

$\displaystyle \text{The equation of tangent is}$

$\displaystyle y-b=-\frac{b}{a}(x-a).$

$\displaystyle ay-ab=-bx+ab.$

$\displaystyle bx+ay=2ab.$

$\displaystyle \frac{x}{a}+\frac{y}{b}=2.$

$\displaystyle \text{So, the given line touches the given curve at the given point.}$

$\displaystyle \textbf{Question 19: }~\text{Find the equation of the tangent to the curve }x=\sin 3t, \\ \;y=\cos 2t\text{ at }t=\frac{\pi}{4}. \hspace{5.0cm} \;[\text{CBSE 2008}]$
$\displaystyle \text{Answer:}$

$\displaystyle x=\sin 3t \text{ and } y=\cos 2t.$

$\displaystyle \frac{dx}{dt}=3\cos 3t \text{ and } \frac{dy}{dt}=-2\sin 2t.$

$\displaystyle \therefore \frac{dy}{dx} =\frac{\frac{dy}{dt}}{\frac{dx}{dt}} =\frac{-2\sin 2t}{3\cos 3t}.$

$\displaystyle \text{Slope of tangent, }m =\left(\frac{dy}{dx}\right)_{t=\frac{\pi}{4}} =\frac{-2\sin\left(\frac{\pi}{2}\right)}{3\cos\left(\frac{3\pi}{4}\right)} =\frac{-2}{-\frac{3}{\sqrt{2}}} =\frac{2\sqrt{2}}{3}.$

$\displaystyle x_1=\sin\left(3\times\frac{\pi}{4}\right) =\sin\left(\frac{3\pi}{4}\right)=\frac{1}{\sqrt{2}} \text{ and } y_1=\cos\left(2\times\frac{\pi}{4}\right) =\cos\left(\frac{\pi}{2}\right)=0.$

$\displaystyle \therefore (x_1,y_1)=\left(\frac{1}{\sqrt{2}},0\right).$

$\displaystyle \text{Equation of tangent is,}$

$\displaystyle y-y_1=m(x-x_1).$

$\displaystyle y-0=\frac{2\sqrt{2}}{3}\left(x-\frac{1}{\sqrt{2}}\right).$

$\displaystyle 3y=2\sqrt{2}x-2.$

$\displaystyle 2\sqrt{2}x-3y-2=0.$

$\displaystyle \textbf{Question 20: }~\text{At what points will the tangents to the curve } \\ y=2x^3-15x^2+36x-21\text{ be parallel to }x\text{-axis? Also, find the equations } \\ \text{of the tangents to the curve at these points.}\hspace{5.0cm} \;[\text{CBSE 2011}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Slope of the }x\text{-axis is }0.$

$\displaystyle \text{Let }(x_1,y_1)\text{ be the required point.}$

$\displaystyle y=2x^3-15x^2+36x-21.$

$\displaystyle \text{Since }(x_1,y_1)\text{ lies on the curve, therefore}$

$\displaystyle y_1=2x_1^3-15x_1^2+36x_1-21\quad\text{...(1)}$

$\displaystyle \text{Now, }y=2x^3-15x^2+36x-21.$

$\displaystyle \frac{dy}{dx}=6x^2-30x+36.$

$\displaystyle \text{Slope of tangent at }(x_1,y_1) =\left(\frac{dy}{dx}\right)_{(x_1,y_1)} =6x_1^2-30x_1+36.$

$\displaystyle \text{Given that slope of tangent = slope of the }x\text{-axis.}$

$\displaystyle 6x_1^2-30x_1+36=0.$

$\displaystyle x_1^2-5x_1+6=0.$

$\displaystyle (x_1-2)(x_1-3)=0.$

$\displaystyle x_1=2 \text{ or } x_1=3.$

$\displaystyle \text{Case 1: }x_1=2.$

$\displaystyle y_1=16-60+72-21=7\quad\text{[From (1)]}.$

$\displaystyle \therefore (x_1,y_1)=(2,7).$

$\displaystyle \text{Equation of tangent is,}$

$\displaystyle y-y_1=m(x-x_1).$

$\displaystyle y-7=0(x-2).$

$\displaystyle y=7.$

$\displaystyle \text{Case 2: }x_1=3.$

$\displaystyle y_1=54-135+108-21=6\quad\text{[From (1)]}.$

$\displaystyle \therefore (x_1,y_1)=(3,6).$

$\displaystyle \text{Equation of tangent is,}$

$\displaystyle y-y_1=m(x-x_1).$

$\displaystyle y-6=0(x-3).$

$\displaystyle y=6.$

$\displaystyle \textbf{Question 21: }~\text{Find the equation of the tangents to the curve } \\ 3x^2-y^2=8,\text{ which passes through the point }\left(\frac{4}{3},0\right).\hspace{5.0cm} \;[\text{CBSE 2013}]$
$\displaystyle \text{Answer:}$