Class 12: Tanget and Normals

$\displaystyle \textbf{Question 1: }~\text{Find the slopes of the tangent and the normal to the following} \\ \text{curves at the indicated points:}$
$\displaystyle (i)\;y=\sqrt{x^3}\text{ at }x=4$

$\displaystyle \text{Answer:}$

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

$\displaystyle \Rightarrow \frac{dy}{dx}=\frac{3}{2}x^{\frac{1}{2}}=\frac{3}{2}\sqrt{x}$

$\displaystyle \text{When } x=4,$

$\displaystyle y=\sqrt{4^3}=\sqrt{64}=8$

$\displaystyle \text{Now,}$

$\displaystyle \text{Slope of the tangent at }(4,8) =\left(\frac{dy}{dx}\right)_{(4,8)} =\frac{3}{2}\sqrt{4}=3$

$\displaystyle \text{Slope of the normal at }(4,8) =-\frac{1}{\left(\frac{dy}{dx}\right)_{(4,8)}} =-\frac{1}{3}$

$\displaystyle (ii)\;y=\sqrt{x}\text{ at }x=9$

$\displaystyle \text{Answer:}$

$\displaystyle y=\sqrt{x}=x^{\frac{1}{2}}$

$\displaystyle \Rightarrow \frac{dy}{dx}=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}$

$\displaystyle \text{When } x=9,$

$\displaystyle y=\sqrt{9}=3$

$\displaystyle \text{Now,}$

$\displaystyle \text{Slope of the tangent at }(9,3) =\left(\frac{dy}{dx}\right)_{(9,3)} =\frac{1}{2\sqrt{9}} =\frac{1}{6}$

$\displaystyle \text{Slope of the normal at }(9,3) =-\frac{1}{\left(\frac{dy}{dx}\right)_{(9,3)}} =-6$

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

$\displaystyle \text{Answer:}$

$\displaystyle y=x^3-x$

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

$\displaystyle \text{When } x=2,$

$\displaystyle y=2^3-2=8-2=6$

$\displaystyle \text{Now,}$

$\displaystyle \text{Slope of the tangent at }(2,6) =\left(\frac{dy}{dx}\right)_{(2,6)} =3(2)^2-1 =11$

$\displaystyle \text{Slope of the normal at }(2,6) =-\frac{1}{\left(\frac{dy}{dx}\right)_{(2,6)}} =-\frac{1}{11}$

$\displaystyle (iv)\;y=2x^2+3\sin x\text{ at }x=0$

$\displaystyle \text{Answer:}$

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

$\displaystyle \Rightarrow \frac{dy}{dx}=4x+3\cos x$

$\displaystyle \text{When } x=0,$

$\displaystyle y=2(0)^2+3\sin 0=0$

$\displaystyle \text{Now,}$

$\displaystyle \text{Slope of the tangent at }(0,0) =\left(\frac{dy}{dx}\right)_{(0,0)} =4(0)+3\cos 0 =3$

$\displaystyle \text{Slope of the normal at }(0,0) =-\frac{1}{\left(\frac{dy}{dx}\right)_{(0,0)}} =-\frac{1}{3}$

$\displaystyle (v)\;x=a(\theta-\sin\theta),\;y=a(1+\cos\theta)\text{ at }\theta=-\frac{\pi}{2}$

$\displaystyle \text{Answer:}$

$\displaystyle x=a(\theta-\sin\theta)$

$\displaystyle \Rightarrow \frac{dx}{d\theta}=a(1-\cos\theta)$

$\displaystyle y=a(1+\cos\theta)$

$\displaystyle \Rightarrow \frac{dy}{d\theta}=-a\sin\theta$

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

$\displaystyle \text{Now,}$

$\displaystyle \text{Slope of the tangent at }\theta=-\frac{\pi}{2} =\left(\frac{dy}{dx}\right)_{\theta=-\frac{\pi}{2}} =-\cot\!\left(-\frac{\pi}{4}\right) =1$

$\displaystyle \text{Slope of the normal} =-\frac{1}{\left(\frac{dy}{dx}\right)_{\theta=-\frac{\pi}{2}}} =-1$

$\displaystyle (vi)\;x=a\cos^3\theta,\;y=a\sin^3\theta\text{ at }\theta=\frac{\pi}{4}$

$\displaystyle \text{Answer:}$

$\displaystyle x=a\cos^3\theta$

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

$\displaystyle y=a\sin^3\theta$

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

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

$\displaystyle \text{Now,}$

$\displaystyle \text{Slope of the tangent at }\theta=\frac{\pi}{4} =\left(\frac{dy}{dx}\right)_{\theta=\frac{\pi}{4}} =-\tan\frac{\pi}{4} =-1$

$\displaystyle \text{Slope of the normal} =-\frac{1}{\left(\frac{dy}{dx}\right)_{\theta=\frac{\pi}{4}}} =1$

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

$\displaystyle \text{Answer:}$

$\displaystyle x=a(\theta-\sin\theta)$

$\displaystyle \Rightarrow \frac{dx}{d\theta}=a(1-\cos\theta)$

$\displaystyle y=a(1-\cos\theta)$

$\displaystyle \Rightarrow \frac{dy}{d\theta}=a\sin\theta$

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

$\displaystyle \text{Now,}$

$\displaystyle \text{Slope of the tangent at }\theta=\frac{\pi}{2} =\left(\frac{dy}{dx}\right)_{\theta=\frac{\pi}{2}} =\cot\!\left(\frac{\pi}{4}\right) =1$

$\displaystyle \text{Slope of the normal} =-\frac{1}{\left(\frac{dy}{dx}\right)_{\theta=\frac{\pi}{2}}} =-1$

$\displaystyle (viii)\;y=(\sin 2x+\cot x+2)^2\text{ at }x=\frac{\pi}{2}$

$\displaystyle \text{Answer:}$

$\displaystyle y=(\sin 2x+\cot x+2)^2$

$\displaystyle \Rightarrow \frac{dy}{dx} =2(\sin 2x+\cot x+2)(2\cos 2x-\mathrm{cosec}^2 x)$

$\displaystyle \text{Now,}$

$\displaystyle \text{Slope of the tangent at }x=\frac{\pi}{2} =\left(\frac{dy}{dx}\right)_{x=\frac{\pi}{2}}$

$\displaystyle =2\!\left[\sin\!\left(2\cdot\frac{\pi}{2}\right) +\cot\!\left(\frac{\pi}{2}\right)+2\right] \!\left[2\cos\!\left(2\cdot\frac{\pi}{2}\right) -\mathrm{cosec}^2\!\left(\frac{\pi}{2}\right)\right]$

$\displaystyle =2(0+0+2)\,(-2-1)$

$\displaystyle =-12$

$\displaystyle \text{Slope of the normal} =-\frac{1}{\left(\frac{dy}{dx}\right)_{x=\frac{\pi}{2}}} =\frac{1}{12}$

$\displaystyle (ix)\;x^2+3y+y^2=5\text{ at }(1,1)$

$\displaystyle \text{Answer:}$

$\displaystyle x^2+3y+y^2=5$

$\displaystyle \text{On differentiating both sides w.r.t. }x,\text{ we get}$

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

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

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

$\displaystyle \text{Now,}$

$\displaystyle \text{Slope of the tangent at }(1,1) =\left(\frac{dy}{dx}\right)_{(1,1)} =\frac{-2(1)}{3+2(1)} =-\frac{2}{5}$

$\displaystyle \text{Slope of the normal at }(1,1) =-\frac{1}{\left(\frac{dy}{dx}\right)_{(1,1)}} =\frac{5}{2}$

$\displaystyle (x)\;xy=6\text{ at }(1,6)$

$\displaystyle \text{Answer:}$

$\displaystyle xy=6$

$\displaystyle \text{On differentiating both sides w.r.t. }x,\text{ we get}$

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

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

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

$\displaystyle \text{Now,}$

$\displaystyle \text{Slope of the tangent at }(1,6) =\left(\frac{dy}{dx}\right)_{(1,6)} =-\frac{6}{1} =-6$

$\displaystyle \text{Slope of the normal at }(1,6) =-\frac{1}{\left(\frac{dy}{dx}\right)_{(1,6)}} =\frac{1}{6}$

$\displaystyle \textbf{Question 2: }~\text{Find the values of }a\text{ and }b\text{ if the slope of the tangent to the curve } \\ xy+ax+by=2\text{ at }(1,1)\text{ is }2.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Given: } xy+ax+by=2 \qquad (1)$

$\displaystyle \text{On differentiating both sides w.r.t. }x,\text{ we get}$

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

$\displaystyle \Rightarrow \frac{dy}{dx}(x+b)=-a-y$

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

$\displaystyle \text{Now,}$

$\displaystyle \left(\frac{dy}{dx}\right)_{(1,1)}=2$

$\displaystyle \Rightarrow \frac{-a-1}{1+b}=2$

$\displaystyle \Rightarrow -a-1=2+2b$

$\displaystyle \Rightarrow -a=3+2b$

$\displaystyle \Rightarrow a=-(3+2b)$

$\displaystyle \text{On substituting }a=-(3+2b),\ x=1,\ y=1\text{ in eq. (1), we get}$

$\displaystyle 1+a+b=2$

$\displaystyle \Rightarrow 1-(3+2b)+b=2$

$\displaystyle \Rightarrow -2-b=2$

$\displaystyle \Rightarrow b=-4$

$\displaystyle \text{and } a=-(3+2b)=-(3-8)=5$

$\displaystyle \therefore a=5 \text{ and } b=-4$

$\displaystyle \textbf{Question 3: }~\text{If the tangent to the curve }y=x^3+ax+b\text{ at }(1,-6)\text{ is parallel to the line } \\ x-y+5=0,\text{ find }a\text{ and }b. \hspace{7.0cm} \;[\text{CBSE 2005}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Given: } x-y+5=0$

$\displaystyle \Rightarrow y=x+5$

$\displaystyle \Rightarrow \frac{dy}{dx}=1$

$\displaystyle \text{Now,}$

$\displaystyle y=x^3+ax+b \qquad (1)$

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

$\displaystyle \text{Slope of the tangent at }(1,-6) =\text{ slope of the given line}$

$\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{(1,-6)}=1$

$\displaystyle \Rightarrow 3(1)^2+a=1$

$\displaystyle \Rightarrow a=-2$

$\displaystyle \text{On substituting }a=-2,\ x=1,\ y=-6\text{ in eq. (1), we get}$

$\displaystyle -6=1-2+b$

$\displaystyle \Rightarrow b=-5$

$\displaystyle \therefore a=-2 \text{ and } b=-5$

$\displaystyle \textbf{Question 4: }~\text{Find a point on the curve }y=x^3-3x\text{ where the tangent is parallel} \\ \text{to the chord joining } (1,-2)\text{ and }(2,2).$
$\displaystyle \text{Answer:}$

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

$\displaystyle \text{Slope of the chord } =\frac{y_2-y_1}{x_2-x_1} =\frac{2+2}{2-1} =4$

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

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

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

$\displaystyle \text{It is given that the tangent and the chord are parallel.}$

$\displaystyle \therefore \text{Slope of the tangent }=\text{ Slope of the chord}$

$\displaystyle \Rightarrow 3x_1^2-3=4$

$\displaystyle \Rightarrow 3x_1^2=7$

$\displaystyle \Rightarrow x_1^2=\frac{7}{3}$

$\displaystyle \Rightarrow x_1=\pm\sqrt{\frac{7}{3}}$

$\displaystyle \text{Case 1: When }x_1=\sqrt{\frac{7}{3}}$

$\displaystyle \text{On substituting this in eq. (1), we get}$

$\displaystyle y_1=\left(\sqrt{\frac{7}{3}}\right)^3-3\sqrt{\frac{7}{3}} =\frac{7}{3}\sqrt{\frac{7}{3}}-3\sqrt{\frac{7}{3}} =-\frac{2}{3}\sqrt{\frac{7}{3}}$

$\displaystyle \therefore (x_1,y_1)=\left(\sqrt{\frac{7}{3}},-\frac{2}{3}\sqrt{\frac{7}{3}}\right)$

$\displaystyle \text{Case 2: When }x_1=-\sqrt{\frac{7}{3}}$

$\displaystyle \text{On substituting this in eq. (1), we get}$

$\displaystyle y_1=\left(-\sqrt{\frac{7}{3}}\right)^3-3\left(-\sqrt{\frac{7}{3}}\right) =-\frac{7}{3}\sqrt{\frac{7}{3}}+3\sqrt{\frac{7}{3}} =\frac{2}{3}\sqrt{\frac{7}{3}}$

$\displaystyle \therefore (x_1,y_1)=\left(-\sqrt{\frac{7}{3}},\frac{2}{3}\sqrt{\frac{7}{3}}\right)$

$\displaystyle \textbf{Question 5: }~\text{Find the points on the curve }y=x^3-2x^2-2x\text{ at which the tangent lines} \\ \text{are parallel to the line }y=2x-3.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Given: } y=2x-3$

$\displaystyle \therefore \text{Slope of the line }=\frac{dy}{dx}=2$

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

$\displaystyle \text{Since }(x_1,y_1)\text{ lies on the curve,} \quad y_1=x_1^3-2x_1^2-2x_1 \qquad (1)$

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

$\displaystyle \text{Slope of the tangent at }(x_1,y_1) =\left(\frac{dy}{dx}\right)_{(x_1,y_1)} =3x_1^2-4x_1-2$

$\displaystyle \text{It is given that the tangent and the given line are parallel.}$

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

$\displaystyle \Rightarrow 3x_1^2-4x_1-2=2$

$\displaystyle \Rightarrow 3x_1^2-4x_1-4=0$

$\displaystyle \Rightarrow 3x_1^2-6x_1+2x_1-4=0$

$\displaystyle \Rightarrow 3x_1(x_1-2)+2(x_1-2)=0$

$\displaystyle \Rightarrow (x_1-2)(3x_1+2)=0$

$\displaystyle \Rightarrow x_1=2 \text{ or } x_1=-\frac{2}{3}$

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

$\displaystyle \text{On substituting this in eq. (1), we get}$

$\displaystyle y_1=2^3-2(2)^2-2(2)=8-8-4=-4$

$\displaystyle \therefore (x_1,y_1)=(2,-4)$

$\displaystyle \text{Case 2: When }x_1=-\frac{2}{3}$

$\displaystyle \text{On substituting this in eq. (1), we get}$

$\displaystyle y_1=\left(-\frac{2}{3}\right)^3 -2\left(-\frac{2}{3}\right)^2 -2\left(-\frac{2}{3}\right) =-\frac{8}{27}-\frac{8}{9}+\frac{4}{3} =\frac{4}{27}$

$\displaystyle \therefore (x_1,y_1)=\left(-\frac{2}{3},\frac{4}{27}\right)$

$\displaystyle \textbf{Question 6: }~\text{Find the points on the curve }y^2=2x^3\text{ at which the slope of the tangent is }3.$
$\displaystyle \text{Answer:}$

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

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

$\displaystyle \text{Since }(x_1,y_1)\text{ lies on the curve,} \quad y_1^2=2x_1^3 \qquad (1)$

$\displaystyle \text{On differentiating both sides w.r.t. }x,\text{ we get}$

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

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

$\displaystyle \text{Slope of the tangent at }(x_1,y_1) =\frac{3x_1^2}{y_1}$

$\displaystyle \text{It is given that the slope of the tangent }=3$

$\displaystyle \therefore \frac{3x_1^2}{y_1}=3 \qquad (2)$

$\displaystyle \Rightarrow y_1=x_1^2$

$\displaystyle \text{On substituting this value of }y_1\text{ in eq. (1), we get}$

$\displaystyle x_1^4=2x_1^3$

$\displaystyle \Rightarrow x_1^3(x_1-2)=0$

$\displaystyle \Rightarrow x_1=0 \text{ or } x_1=2$

$\displaystyle \text{Case 1: When }x_1=0,\ y_1=x_1^2=0$

$\displaystyle \text{The point obtained is }(0,0),\text{ but it does not satisfy eq. (2).}$

$\displaystyle \text{Hence, }(0,0)\text{ is rejected.}$

$\displaystyle \text{Case 2: When }x_1=2,\ y_1=x_1^2=4$

$\displaystyle \therefore \text{The required point is }(2,4)$

$\displaystyle \textbf{Question 7: }~\text{Find the points on the curve }xy+4=0\text{ at which the tangents } \\ \text{are inclined at an angle of }45^\circ\text{ with the }x\text{-axis.}$
$\displaystyle \text{Answer:}$

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

$\displaystyle \text{Slope of the tangent at this point }=\tan 45^\circ=1$

$\displaystyle \text{Given: } xy+4=0 \qquad (1)$

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

$\displaystyle x_1y_1+4=0 \qquad (2)$

$\displaystyle \text{On differentiating eq. (1) w.r.t. }x,\text{ we get}$

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

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

$\displaystyle \text{Slope of the tangent at }(x_1,y_1) =\left(\frac{dy}{dx}\right)_{(x_1,y_1)} =-\frac{y_1}{x_1}$

$\displaystyle \text{Given slope of the tangent }=1$

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

$\displaystyle \Rightarrow x_1=-y_1$

$\displaystyle \text{On substituting }x_1=-y_1\text{ in eq. (2), we get}$

$\displaystyle -\,y_1^2+4=0$

$\displaystyle \Rightarrow y_1^2=4$

$\displaystyle \Rightarrow y_1=\pm 2$

$\displaystyle \text{Case 1: When }y_1=2,\ x_1=-2$

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

$\displaystyle \text{Case 2: When }y_1=-2,\ x_1=2$

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

$\displaystyle \textbf{Question 8: }~\text{Find the point on the curve }y=x^2\text{ where the slope of the } \\ \text{tangent is equal to the }x\text{-coordinate of the point.}$
$\displaystyle \text{Answer:}$

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

$\displaystyle \text{Given: } y=x^2$

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

$\displaystyle y_1=x_1^2 \qquad (1)$

$\displaystyle \text{Now,}$

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

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

$\displaystyle \text{It is given that the slope of the tangent equals the }x\text{-coordinate of the point.}$

$\displaystyle \therefore 2x_1=x_1$

$\displaystyle \Rightarrow x_1=0$

$\displaystyle \text{On substituting }x_1=0\text{ in eq. (1), we get}$

$\displaystyle y_1=0^2=0$

$\displaystyle \therefore \text{The required point is }(0,0)$

$\displaystyle \textbf{Question 9: }~\text{At what points on the circle }x^2+y^2-2x-4y+1=0, \\ \text{ the tangent is parallel to }x\text{-axis.} \hspace{7.0cm} \;[\text{CBSE 2002C}]$
$\displaystyle \text{Answer:}$

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

$\displaystyle \text{We know that the slope of the }x\text{-axis is }0.$

$\displaystyle \text{Given: } x^2+y^2-2x-4y+1=0$

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

$\displaystyle x_1^2+y_1^2-2x_1-4y_1+1=0 \qquad (1)$

$\displaystyle \text{Now, differentiating }x^2+y^2-2x-4y+1=0\text{ w.r.t. }x,$

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

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

$\displaystyle \Rightarrow \frac{dy}{dx}=\frac{2-2x}{2y-4} =\frac{1-x}{y-2}$

$\displaystyle \text{Slope of the tangent at }(x_1,y_1) =\left(\frac{dy}{dx}\right)_{(x_1,y_1)} =\frac{1-x_1}{y_1-2} \qquad (2)$

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

$\displaystyle \therefore \frac{1-x_1}{y_1-2}=0$

$\displaystyle \Rightarrow 1-x_1=0$

$\displaystyle \Rightarrow x_1=1$

$\displaystyle \text{On substituting }x_1=1\text{ in eq. (1), we get}$

$\displaystyle 1+y_1^2-2-4y_1+1=0$

$\displaystyle \Rightarrow y_1^2-4y_1=0$

$\displaystyle \Rightarrow y_1(y_1-4)=0$

$\displaystyle \Rightarrow y_1=0 \text{ or } y_1=4$

$\displaystyle \therefore \text{The required points are }(1,0)\text{ and }(1,4)$

$\displaystyle \textbf{Question 10: }~\text{At what point of the curve }y=x^2\text{ does the tangent } \\ \text{make an angle of }45^\circ\text{ with the }x\text{-axis?}$
$\displaystyle \text{Answer:}$

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

$\displaystyle \text{The tangent makes an angle }45^\circ\text{ with the }x\text{-axis.}$

$\displaystyle \therefore \text{Slope of the tangent }=\tan 45^\circ=1$

$\displaystyle \text{Since the point lies on the curve }y^2=x,$

$\displaystyle y_1^2=x_1 \qquad (1)$

$\displaystyle \text{Now, } y^2=x$

$\displaystyle \Rightarrow 2y\frac{dy}{dx}=1$

$\displaystyle \Rightarrow \frac{dy}{dx}=\frac{1}{2y}$

$\displaystyle \text{Slope of the tangent at }(x_1,y_1) =\left(\frac{dy}{dx}\right)_{(x_1,y_1)} =\frac{1}{2y_1}$

$\displaystyle \text{Given slope of the tangent }=1$

$\displaystyle \Rightarrow \frac{1}{2y_1}=1$

$\displaystyle \Rightarrow 2y_1=1$

$\displaystyle \Rightarrow y_1=\frac{1}{2}$

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

$\displaystyle x_1=y_1^2=\left(\frac{1}{2}\right)^2=\frac{1}{4}$

$\displaystyle \therefore (x_1,y_1)=\left(\frac{1}{4},\frac{1}{2}\right)$

$\displaystyle \textbf{Question 11: }~\text{Find the points on the curve }y=3x^2-9x+8\text{ at which the tangents } \\ \text{are equally inclined with the axes.}$
$\displaystyle \text{Answer:}$

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

$\displaystyle \text{It is given that the tangent at this point is equally inclined to the axes.}$

$\displaystyle \text{Hence, the angle made by the tangent with the }x\text{-axis is } \pm 45^\circ.$

$\displaystyle \therefore \text{Slope of the tangent }=\tan(\pm 45^\circ)=\pm 1 \qquad (1)$

$\displaystyle \text{Since the point lies on the curve }y=3x^2-9x+8,$

$\displaystyle y_1=3x_1^2-9x_1+8$

$\displaystyle \text{Now, } y=3x^2-9x+8$

$\displaystyle \Rightarrow \frac{dy}{dx}=6x-9$

$\displaystyle \text{Slope of the tangent at }(x_1,y_1) =\left(\frac{dy}{dx}\right)_{(x_1,y_1)} =6x_1-9 \qquad (2)$

$\displaystyle \text{From eq. (1) and eq. (2), we get}$

$\displaystyle 6x_1-9=\pm 1$

$\displaystyle \Rightarrow 6x_1-9=1 \text{ or } 6x_1-9=-1$

$\displaystyle \Rightarrow 6x_1=10 \text{ or } 6x_1=8$

$\displaystyle \Rightarrow x_1=\frac{5}{3} \text{ or } x_1=\frac{4}{3}$

$\displaystyle \text{Now, } y_1=3x_1^2-9x_1+8$

$\displaystyle \text{For }x_1=\frac{5}{3},\quad y_1=3\left(\frac{5}{3}\right)^2-9\left(\frac{5}{3}\right)+8 =\frac{25}{3}-15+8=\frac{4}{3}$

$\displaystyle \text{For }x_1=\frac{4}{3},\quad y_1=3\left(\frac{4}{3}\right)^2-9\left(\frac{4}{3}\right)+8 =\frac{16}{3}-12+8=\frac{4}{3}$

$\displaystyle \therefore \text{The required points are } \left(\frac{5}{3},\frac{4}{3}\right) \text{ and } \left(\frac{4}{3},\frac{4}{3}\right)$

$\displaystyle \textbf{Question 12: }~\text{At what points on the curve }y=2x^2-x+1\text{ is the } \\ \text{tangent parallel to the line }y=3x+4?$
$\displaystyle \text{Answer:}$

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

$\displaystyle \text{The slope of the line }y=3x+4\text{ is }3.$

$\displaystyle \text{Since the point lies on the curve }y=2x^2-x+1,$

$\displaystyle y_1=2x_1^2-x_1+1 \qquad (1)$

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

$\displaystyle \Rightarrow \frac{dy}{dx}=4x-1$

$\displaystyle \text{Slope of the tangent at }(x_1,y_1) =\left(\frac{dy}{dx}\right)_{(x_1,y_1)} =4x_1-1$

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

$\displaystyle \Rightarrow 4x_1-1=3$

$\displaystyle \Rightarrow 4x_1=4$

$\displaystyle \Rightarrow x_1=1$

$\displaystyle \text{On substituting }x_1=1\text{ in eq. (1), we get}$

$\displaystyle y_1=2(1)^2-1+1=2$

$\displaystyle \therefore \text{The required point is }(1,2)$

$\displaystyle \textbf{Question 13: }~\text{Find the point on the curve }y=3x^2+4\text{ at which the tangent } \\ \text{is perpendicular to the line whose slope is }-\frac{1}{6}.$
$\displaystyle \text{Answer:}$

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

$\displaystyle \text{Slope of the given line }=-\frac{1}{6}.$

$\displaystyle \therefore \text{Slope of the line perpendicular to it }=6.$

$\displaystyle \text{Since the point lies on the curve }y=3x^2+4,$

$\displaystyle y_1=3x_1^2+4 \qquad (1)$

$\displaystyle \text{Now, } y=3x^2+4$

$\displaystyle \Rightarrow \frac{dy}{dx}=6x$

$\displaystyle \text{Slope of the tangent at }(x_1,y_1) =\left(\frac{dy}{dx}\right)_{(x_1,y_1)} =6x_1$

$\displaystyle \text{Given that the tangent is perpendicular to the given line.}$

$\displaystyle \Rightarrow 6x_1=6$

$\displaystyle \Rightarrow x_1=1$

$\displaystyle \text{On substituting }x_1=1\text{ in eq. (1), we get}$

$\displaystyle y_1=3(1)^2+4=7$

$\displaystyle \therefore \text{The required point is }(1,7)$

$\displaystyle \textbf{Question 14: }~\text{Find the points on the curve }x^2+y^2=13,\text{ the tangent } \\ \text{at each one of which is parallel to the line }2x+3y=7.$
$\displaystyle \text{Answer:}$

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

$\displaystyle \text{The slope of the line }2x+3y=7\text{ is }-\frac{2}{3}.$

$\displaystyle \text{Since the point lies on the curve }x^2+y^2=13,$

$\displaystyle x_1^2+y_1^2=13 \qquad (1)$

$\displaystyle \text{Now, } x^2+y^2=13$

$\displaystyle \text{On differentiating both sides w.r.t. }x,\text{ we get}$

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

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

$\displaystyle \text{Slope of the tangent at }(x_1,y_1) =\left(\frac{dy}{dx}\right)_{(x_1,y_1)} =-\frac{x_1}{y_1}$

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

$\displaystyle \Rightarrow -\frac{x_1}{y_1}=-\frac{2}{3}$

$\displaystyle \Rightarrow x_1=\frac{2y_1}{3} \qquad (2)$

$\displaystyle \text{From eq. (1), we get}$

$\displaystyle \left(\frac{2y_1}{3}\right)^2+y_1^2=13$

$\displaystyle \Rightarrow \frac{4y_1^2}{9}+y_1^2=13$

$\displaystyle \Rightarrow \frac{13y_1^2}{9}=13$

$\displaystyle \Rightarrow y_1^2=9$

$\displaystyle \Rightarrow y_1=\pm 3$

$\displaystyle \text{When }y_1=3,\quad x_1=\frac{2(3)}{3}=2$

$\displaystyle \text{When }y_1=-3,\quad x_1=\frac{2(-3)}{3}=-2$

$\displaystyle \therefore \text{The required points are }(2,3)\text{ and }(-2,-3)$

$\displaystyle \textbf{Question 15: }~\text{Find the points on the curve }2a^2y=x^3-3ax^2\text{ where the tangent } \\ \text{is parallel to }x\text{-axis.}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let }(x_1,y_1)\text{ represent the required points.}$

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

$\displaystyle \text{Given: } 2a^2y=x^3-3ax^2$

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

$\displaystyle 2a^2y_1=x_1^3-3ax_1^2 \qquad (1)$

$\displaystyle \text{Now, differentiating }2a^2y=x^3-3ax^2\text{ w.r.t. }x,$

$\displaystyle 2a^2\frac{dy}{dx}=3x^2-6ax$

$\displaystyle \Rightarrow \frac{dy}{dx}=\frac{3x^2-6ax}{2a^2}$

$\displaystyle \text{Slope of the tangent at }(x_1,y_1) =\left(\frac{dy}{dx}\right)_{(x_1,y_1)} =\frac{3x_1^2-6ax_1}{2a^2}$

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

$\displaystyle \Rightarrow \frac{3x_1^2-6ax_1}{2a^2}=0$

$\displaystyle \Rightarrow 3x_1^2-6ax_1=0$

$\displaystyle \Rightarrow x_1(3x_1-6a)=0$

$\displaystyle \Rightarrow x_1=0 \text{ or } x_1=2a$

$\displaystyle \text{Case 1: When }x_1=0$

$\displaystyle \text{From eq. (1), }2a^2y_1=0$

$\displaystyle \Rightarrow y_1=0$

$\displaystyle \text{Case 2: When }x_1=2a$

$\displaystyle \text{From eq. (1), }2a^2y_1=(2a)^3-3a(2a)^2$

$\displaystyle \Rightarrow 2a^2y_1=8a^3-12a^3=-4a^3$

$\displaystyle \Rightarrow y_1=-2a$

$\displaystyle \therefore \text{The required points are }(0,0)\text{ and }(2a,-2a)$

$\displaystyle \textbf{Question 16: }~\text{At what points on the curve }y=x^2-4x+5\text{ is the tangent } \\ \text{perpendicular to the line }2y+x=7?$
$\displaystyle \text{Answer:}$

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

$\displaystyle \text{Slope of the given line }=-\frac{1}{2}.$

$\displaystyle \therefore \text{Slope of the line perpendicular to it }=2.$

$\displaystyle \text{Since the point lies on the curve }y=x^2-4x+5,$

$\displaystyle y_1=x_1^2-4x_1+5 \qquad (1)$

$\displaystyle \text{Now, } y=x^2-4x+5$

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

$\displaystyle \text{Slope of the tangent at }(x_1,y_1) =\left(\frac{dy}{dx}\right)_{(x_1,y_1)} =2x_1-4$

$\displaystyle \text{Given that the tangent is perpendicular to the given line.}$

$\displaystyle \Rightarrow 2x_1-4=2$

$\displaystyle \Rightarrow 2x_1=6$

$\displaystyle \Rightarrow x_1=3$

$\displaystyle \text{On substituting }x_1=3\text{ in eq. (1), we get}$

$\displaystyle y_1=3^2-4(3)+5=9-12+5=2$

$\displaystyle \therefore \text{The required point is }(3,2)$

$\displaystyle \textbf{Question 17: }~\text{Find the points on the curve }\frac{x^2}{4}+\frac{y^2}{25}=1\text{ at which } \\ \text{the tangents are parallel to }(i)\;x\text{-axis }\;(ii)\;y\text{-axis.}$

$\displaystyle \text{Answer:}$
$\displaystyle \text{(i)}$

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

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

$\displaystyle \text{Since the point lies on the curve } \frac{x^2}{4}+\frac{y^2}{25}=1,$

$\displaystyle \frac{x_1^2}{4}+\frac{y_1^2}{25}=1 \qquad (1)$

$\displaystyle \text{Now, } \frac{x^2}{4}+\frac{y^2}{25}=1$

$\displaystyle \text{On differentiating both sides w.r.t. }x,\text{ we get}$

$\displaystyle \frac{2x}{4}+\frac{2y}{25}\frac{dy}{dx}=0$

$\displaystyle \Rightarrow \frac{2y}{25}\frac{dy}{dx}=-\frac{x}{2}$

$\displaystyle \Rightarrow \frac{dy}{dx}=-\frac{25x}{4y}$

$\displaystyle \text{Slope of the tangent at }(x_1,y_1) =\left(\frac{dy}{dx}\right)_{(x_1,y_1)} =-\frac{25x_1}{4y_1}$

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

$\displaystyle \Rightarrow -\frac{25x_1}{4y_1}=0$

$\displaystyle \Rightarrow x_1=0$

$\displaystyle \text{On substituting }x_1=0\text{ in eq. (1), we get}$

$\displaystyle \frac{y_1^2}{25}=1$

$\displaystyle \Rightarrow y_1^2=25$

$\displaystyle \Rightarrow y_1=\pm 5$

$\displaystyle \therefore \text{The required points are }(0,5)\text{ and }(0,-5)$

$\displaystyle \text{(ii)}$

$\displaystyle \text{The slope of the }y\text{-axis is } \infty.$

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

$\displaystyle \text{Since the point lies on the curve } \frac{x^2}{4}+\frac{y^2}{25}=1,$

$\displaystyle \frac{x_1^2}{4}+\frac{y_1^2}{25}=1 \qquad (1)$

$\displaystyle \text{Now, } \frac{x^2}{4}+\frac{y^2}{25}=1$

$\displaystyle \text{On differentiating both sides w.r.t. }x,\text{ we get}$

$\displaystyle \frac{2x}{4}+\frac{2y}{25}\frac{dy}{dx}=0$

$\displaystyle \Rightarrow \frac{2y}{25}\frac{dy}{dx}=-\frac{x}{2}$

$\displaystyle \Rightarrow \frac{dy}{dx}=-\frac{25x}{4y}$

$\displaystyle \text{Slope of the tangent at }(x_1,y_1) =\left(\frac{dy}{dx}\right)_{(x_1,y_1)} =-\frac{25x_1}{4y_1}$

$\displaystyle \text{Given that the slope of the tangent equals the slope of the }y\text{-axis.}$

$\displaystyle \Rightarrow -\frac{25x_1}{4y_1}=\infty$

$\displaystyle \Rightarrow 4y_1=0$

$\displaystyle \Rightarrow y_1=0$

$\displaystyle \text{On substituting }y_1=0\text{ in eq. (1), we get}$

$\displaystyle \frac{x_1^2}{4}=1$

$\displaystyle \Rightarrow x_1^2=4$

$\displaystyle \Rightarrow x_1=\pm 2$

$\displaystyle \therefore \text{The required points are }(2,0)\text{ and }(-2,0)$

$\displaystyle \textbf{Question 18: }~\text{Find the points on the curve }x^2+y^2-2x-3=0\text{ at which the } \\ \text{tangents are parallel to }(i)\;x\text{-axis }\;(ii)\;y\text{-axis.} \hspace{5.0cm} [\text{CBSE 2011}]$
$\displaystyle \text{Answer:}$

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

$\displaystyle \text{Since the point lies on the curve }x^2+y^2-2x-3=0,$

$\displaystyle x_1^2+y_1^2-2x_1-3=0 \qquad (1)$

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

$\displaystyle \text{On differentiating both sides w.r.t. }x,\text{ we get}$

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

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

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

$\displaystyle \text{Slope of the tangent at }(x_1,y_1) =\left(\frac{dy}{dx}\right)_{(x_1,y_1)} =\frac{1-x_1}{y_1}$

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

$\displaystyle \Rightarrow \frac{1-x_1}{y_1}=0$

$\displaystyle \Rightarrow 1-x_1=0$

$\displaystyle \Rightarrow x_1=1$

$\displaystyle \text{From eq. (1), we get}$

$\displaystyle 1+y_1^2-2-3=0$

$\displaystyle \Rightarrow y_1^2-4=0$

$\displaystyle \Rightarrow y_1=\pm 2$

$\displaystyle \therefore \text{The required points are }(1,2)\text{ and }(1,-2)$

$\displaystyle \textbf{Question 19: }~\text{Find the points on the curve }\frac{x^2}{9}+\frac{y^2}{16}=1\text{ at which } \\ \text{the tangents are }(i)\;\text{parallel to }x\text{-axis }\;(ii)\;\text{parallel to }y\text{-axis.}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }$

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

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

$\displaystyle \text{Since the point lies on the curve } \frac{x^2}{9}+\frac{y^2}{16}=1,$

$\displaystyle \frac{x_1^2}{9}+\frac{y_1^2}{16}=1 \qquad (1)$

$\displaystyle \text{Now, } \frac{x^2}{9}+\frac{y^2}{16}=1$

$\displaystyle \text{On differentiating both sides w.r.t. }x,\text{ we get}$

$\displaystyle \frac{2x}{9}+\frac{2y}{16}\frac{dy}{dx}=0$

$\displaystyle \Rightarrow \frac{2y}{16}\frac{dy}{dx}=-\frac{2x}{9}$

$\displaystyle \Rightarrow \frac{dy}{dx}=-\frac{16x}{9y}$

$\displaystyle \text{Slope of the tangent at }(x_1,y_1) =\left(\frac{dy}{dx}\right)_{(x_1,y_1)} =-\frac{16x_1}{9y_1}$

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

$\displaystyle \Rightarrow -\frac{16x_1}{9y_1}=0$

$\displaystyle \Rightarrow x_1=0$

$\displaystyle \text{On substituting }x_1=0\text{ in eq. (1), we get}$

$\displaystyle \frac{y_1^2}{16}=1$

$\displaystyle \Rightarrow y_1^2=16$

$\displaystyle \Rightarrow y_1=\pm 4$

$\displaystyle \therefore \text{The required points are }(0,4)\text{ and }(0,-4)$

$\displaystyle \text{(ii) }$

$\displaystyle \text{The slope of the }y\text{-axis is } \infty.$

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

$\displaystyle \text{Since the point lies on the curve } \frac{x^2}{9}+\frac{y^2}{16}=1,$

$\displaystyle \frac{x_1^2}{9}+\frac{y_1^2}{16}=1 \qquad (1)$

$\displaystyle \text{Now, } \frac{x^2}{9}+\frac{y^2}{16}=1$

$\displaystyle \text{On differentiating both sides w.r.t. }x,\text{ we get}$

$\displaystyle \frac{2x}{9}+\frac{2y}{16}\frac{dy}{dx}=0$

$\displaystyle \Rightarrow \frac{2y}{16}\frac{dy}{dx}=-\frac{2x}{9}$

$\displaystyle \Rightarrow \frac{dy}{dx}=-\frac{16x}{9y}$

$\displaystyle \text{Slope of the tangent at }(x_1,y_1) =\left(\frac{dy}{dx}\right)_{(x_1,y_1)} =-\frac{16x_1}{9y_1}$

$\displaystyle \text{Given that the slope of the tangent equals the slope of the }y\text{-axis.}$

$\displaystyle \Rightarrow -\frac{16x_1}{9y_1}=\infty$

$\displaystyle \Rightarrow y_1=0$

$\displaystyle \text{On substituting }y_1=0\text{ in eq. (1), we get}$

$\displaystyle \frac{x_1^2}{9}=1$

$\displaystyle \Rightarrow x_1^2=9$

$\displaystyle \Rightarrow x_1=\pm 3$

$\displaystyle \therefore \text{The required points are }(3,0)\text{ and }(-3,0)$

$\displaystyle \textbf{Question 20: }~\text{Show that the tangents to the curve }y=7x^3+11\text{ at the points }x=2\text{ and }x=-2\text{ are parallel.}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Given: } y=7x^3+11$

$\displaystyle \Rightarrow \frac{dy}{dx}=21x^2$

$\displaystyle \text{Now,}$

$\displaystyle \text{Slope of the tangent at }x=2 =\left(\frac{dy}{dx}\right)_{x=2} =21(2)^2 =84$

$\displaystyle \text{Slope of the tangent at }x=-2 =\left(\frac{dy}{dx}\right)_{x=-2} =21(-2)^2 =84$

$\displaystyle \text{Both slopes are equal. Hence, the tangents at }x=2\text{ and }x=-2\text{ are parallel.}$

$\displaystyle \textbf{Question 21: }~\text{Find the points on the curve }y=x^3\text{ where the slope of the } \\ \text{tangent is equal to }x\text{-coordinate of the point.} \hspace{5.0cm} \;[\text{CBSE 2008}]$
$\displaystyle \text{Answer:}$