\displaystyle \textbf{Question 1. }\text{Find point on } y=(x-2)^2 \text{ where tangent is parallel to} \\ \text{chord joining }(2,0),(4,4)\ \text{ISC 2024} \ \
\displaystyle \text{Answer:}
\displaystyle \text{4. Given, equation of curve is } y=(x-2)^{2} \qquad \cdots(i)
\displaystyle \text{Let } (h,k) \text{ be the point at which the tangent is parallel to the line joining the points } (2,0) \text{ and } (4,4)
\displaystyle \text{Then, slope of tangent at } (h,k) \text{ is } \left(\frac{dy}{dx}\right)_{(h,k)}=2(h-2)
\displaystyle \text{Also, slope of the chord joining the points } (2,0) \text{ and } (4,4) \text{ is } \frac{4-0}{4-2}=2
\displaystyle \text{Since tangent is parallel to chord,}
\displaystyle 2(h-2)=2 \Rightarrow h-2=1 \Rightarrow h=3
\displaystyle \text{Also, point } (h,k) \text{ lies on curve, so}
\displaystyle k=(h-2)^{2}=(3-2)^{2}=1
\displaystyle \text{Hence, the required point is } (3,1).
\\

\displaystyle \textbf{Question 2. }\text{For the curve } y^2=2x^3-7,\text{ the slope of the normal at }(2,3)\text{ is } \hspace{0.2cm} \text{ISC 2023} \ \
\displaystyle \text{(a) }4 \quad (b)\ \frac{1}{4} \quad (c)\ -4 \quad (d)\ -\frac{1}{4} \ \
\displaystyle \text{Answer:}
\displaystyle \text{1. (d) We have, } y^{2}=2x^{3}-7
\displaystyle \text{On differentiating w.r.t. } x, \text{ we get}
\displaystyle 2y\frac{dy}{dx}=6x^{2}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{3x^{2}}{y}
\displaystyle \text{Slope of tangent at } (2,3) \text{ is } \left(\frac{dy}{dx}\right)_{(2,3)}=\frac{3(2)^{2}}{3}=\frac{12}{3}=4
\displaystyle \text{Slope of normal at } (2,3)=-\frac{1}{4}
\\

\displaystyle \textbf{Question 3. }\text{Curve } y^2=px^3+q \text{ tangent at }(2,3)\text{ is } y=4x-7 \ \text{ISC 2020} \ \
\displaystyle \text{Find } p,q \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, curve is } y^{2}=px^{3}+q
\displaystyle \text{Since it passes through } (2,3)
\displaystyle \Rightarrow (3)^{2}=p(2)^{3}+q
\displaystyle \Rightarrow 9=8p+q \qquad \cdots(i)
\displaystyle \text{Now, differentiating } y^{2}=px^{3}+q,
\displaystyle 2y\frac{dy}{dx}=3px^{2}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{3px^{2}}{2y}
\displaystyle \text{At } (2,3), \left(\frac{dy}{dx}\right)=\frac{3p(2)^{2}}{2(3)}=2p
\displaystyle \text{Equation of tangent is } y=4x-7 \Rightarrow \text{slope }=4
\displaystyle \therefore 2p=4 \Rightarrow p=2
\displaystyle \text{Putting in (i), } 8(2)+q=9 \Rightarrow q=9-16=-7
\displaystyle \therefore p=2,\ q=-7
\\

\displaystyle \textbf{Question 4. }\text{Find the points on the curve } y=4x^3-3x+5 \text{ at which} \ \
\displaystyle \text{the equation of the tangent is parallel to the X-axis. ISC 2018} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } y=4x^{3}-3x+5
\displaystyle \frac{dy}{dx}=12x^{2}-3
\displaystyle \text{For tangent parallel to } x\text{-axis, } \frac{dy}{dx}=0
\displaystyle \Rightarrow 12x^{2}-3=0 \Rightarrow x^{2}=\frac{1}{4}
\displaystyle \Rightarrow x=\pm \frac{1}{2}
\displaystyle \text{When } x=\frac{1}{2},\ y=4 \quad \text{and when } x=-\frac{1}{2},\ y=6
\displaystyle \therefore \text{Points are } \left(\frac{1}{2},4\right) \text{ and } \left(-\frac{1}{2},6\right)
\\

\displaystyle \textbf{Question 5. }\text{Find the point on the curve } 9y^2=x^3,\ \text{where the} \ \
\displaystyle \text{normal to the curve makes equal intercepts on the axes.} \ \text{ \hspace{2.2cm} ISC 2017} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } 9y^{2}=x^{3} \qquad \cdots(i)
\displaystyle \text{Differentiating, } 18y\frac{dy}{dx}=3x^{2}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{x^{2}}{6y}
\displaystyle \text{Slope of normal }=-\frac{1}{dy/dx}=-\frac{6y}{x^{2}}
\displaystyle \text{For equal intercepts, slope }=\pm 1
\displaystyle \Rightarrow -\frac{6y_{1}}{x_{1}^{2}}=\pm 1 \Rightarrow y_{1}=\mp \frac{x_{1}^{2}}{6}
\displaystyle \text{Substitute in (i): } 9\left(\frac{x_{1}^{2}}{6}\right)^{2}=x_{1}^{3}
\displaystyle \Rightarrow \frac{9x_{1}^{4}}{36}=x_{1}^{3} \Rightarrow \frac{x_{1}^{4}}{4}=x_{1}^{3}
\displaystyle \Rightarrow x_{1}=4 \ (\because x_{1}\neq 0)
\displaystyle \Rightarrow y_{1}=\pm \frac{16}{6}=\pm \frac{8}{3}
\displaystyle \therefore \text{Points are } (4,\frac{8}{3}) \text{ and } (4,-\frac{8}{3})
\\

\displaystyle \textbf{Question 6. }\text{Find the equations of the tangent and normal to the} \ \
\displaystyle \text{curves } x=a\sin^3\theta \text{ and } y=a\cos^3\theta \text{ at } \theta=\frac{\pi}{4}. \ \text{ \hspace{2.2cm} ISC 2016} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } x=a\sin^{3}\theta \qquad \cdots(i)
\displaystyle \frac{dx}{d\theta}=a\cdot 3\sin^{2}\theta\cos\theta
\displaystyle \Rightarrow \frac{dx}{d\theta}=3a\sin^{2}\theta\cos\theta
\displaystyle y=a\cos^{3}\theta \qquad \cdots(ii)
\displaystyle \frac{dy}{d\theta}=-3a\cos^{2}\theta\sin\theta
\displaystyle \frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{-3a\cos^{2}\theta\sin\theta}{3a\sin^{2}\theta\cos\theta}=-\cot\theta
\displaystyle \text{At } \theta=\frac{\pi}{4},\ \frac{dy}{dx}=-1
\displaystyle x=a\left(\sin\frac{\pi}{4}\right)^{3}=a\left(\frac{1}{\sqrt{2}}\right)^{3}=\frac{a}{(2)^{3/2}}
\displaystyle y=a\left(\cos\frac{\pi}{4}\right)^{3}=\frac{a}{(2)^{3/2}}
\displaystyle \text{Equation of tangent: } y-y_{1}=\frac{dy}{dx}(x-x_{1})
\displaystyle y-\frac{a}{(2)^{3/2}}=-1\left(x-\frac{a}{(2)^{3/2}}\right)
\displaystyle \Rightarrow y+x=\frac{2a}{(2)^{3/2}}
\displaystyle \Rightarrow y+x=\frac{a}{\sqrt{2}}
\displaystyle \Rightarrow x+y-\frac{a}{\sqrt{2}}=0
\displaystyle \text{Clearly, slope of normal }=-\frac{1}{\text{slope of tangent}}
\displaystyle \Rightarrow \text{Slope of normal }=-\frac{1}{-1}=1
\displaystyle \text{Equation of normal at } \left(\frac{a}{2^{3/2}},\frac{a}{2^{3/2}}\right) \text{ is}
\displaystyle y-\frac{a}{2^{3/2}}=1\left(x-\frac{a}{2^{3/2}}\right)
\displaystyle \Rightarrow x-y=0
\\

\displaystyle \textbf{Question 7. }\text{Find the points on the curve } x^2+y^2-2x-3=0 \text{ at} \ \
\displaystyle \text{which tangent is parallel to X-axis.} \ \text{ \hspace{2.2cm} ISC 2015} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, equation of curve is } x^{2}+y^{2}-2x-3=0 \qquad \cdots(i)
\displaystyle \text{On differentiating both sides of Eq. (i) 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{2-2x}{2y}=\frac{1-x}{y}
\displaystyle \text{For tangent parallel to } x\text{-axis, } \frac{dy}{dx}=0
\displaystyle \Rightarrow 1-x=0 \Rightarrow x=1
\displaystyle \text{Putting } x=1 \text{ in Eq. (i), we get}
\displaystyle 1+y^{2}-2-3=0
\displaystyle \Rightarrow y^{2}-4=0 \Rightarrow y=\pm 2
\displaystyle \text{Hence, the required points are } (1,2) \text{ and } (1,-2).
\\

\displaystyle \textbf{Question 8. }\text{Find the points on the curve } y=x^3 \text{ at which the slope} \ \
\displaystyle \text{of the tangent is equal to y-coordinate of the point.} \ \text{ \hspace{2.2cm} ISC 2013} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, equation of curve is } y=x^{3} \qquad \cdots(i)
\displaystyle \text{On differentiating both sides w.r.t. } x, \text{ we get}
\displaystyle \frac{dy}{dx}=3x^{2}
\displaystyle \therefore \text{Slope of tangent at any point } (x,y) \text{ is } \frac{dy}{dx}=3x^{2}
\displaystyle \text{Given that slope of tangent }=y\text{-coordinate of the point}
\displaystyle \Rightarrow \frac{dy}{dx}=y
\displaystyle \Rightarrow 3x^{2}=y
\displaystyle \Rightarrow 3x^{2}=x^{3} \quad [\because y=x^{3}]
\displaystyle \Rightarrow 3x^{2}-x^{3}=0 \Rightarrow x^{2}(3-x)=0
\displaystyle \Rightarrow x=0 \text{ or } x=3
\displaystyle \text{Putting } x=0,3 \text{ in Eq. (i), we get}
\displaystyle y=0^{3}=0 \text{ and } y=3^{3}=27
\displaystyle \text{Hence, the required points are } (0,0) \text{ and } (3,27).
\\

\displaystyle \textbf{Question 9. }\text{Find the equations of tangents to the curve} \ \
\displaystyle y=(x^2-1)(x-2)\ \text{at the points, where the curve cuts the X-axis.} \ \text{ \hspace{2.2cm} ISC 2010} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, equation of curve is } y=(x^{2}-1)(x-2) \qquad \cdots(i)
\displaystyle \text{Since the curve cuts the } x\text{-axis, at that point } y=0
\displaystyle \text{Putting } y=0, \text{ we get } (x^{2}-1)(x-2)=0
\displaystyle \Rightarrow x^{2}=1 \text{ or } x=2
\displaystyle \Rightarrow x=-1,1,2
\displaystyle \text{Thus, the curve cuts the } x\text{-axis at } (-1,0),(1,0),(2,0)
\displaystyle \text{On differentiating Eq. (i) w.r.t. } x, \text{ we get}
\displaystyle \frac{dy}{dx}=(x^{2}-1)\cdot 1+(x-2)\cdot 2x
\displaystyle \text{[by using product rule]}
\displaystyle \Rightarrow \frac{dy}{dx}=x^{2}-1+2x^{2}-4x
\displaystyle \Rightarrow \frac{dy}{dx}=3x^{2}-4x-1
\displaystyle \text{Slope at } (-1,0):\ m_{1}=3(-1)^{2}-4(-1)-1=3+4-1=6
\displaystyle \text{Slope at } (1,0):\ m_{2}=3(1)^{2}-4(1)-1=3-4-1=-2
\displaystyle \text{Slope at } (2,0):\ m_{3}=3(2)^{2}-4(2)-1=12-8-1=3
\displaystyle \text{Equation of tangent is } y-y_{1}=m(x-x_{1})
\displaystyle \text{Equation of tangent at } (-1,0),\ m_{1}=6
\displaystyle y-0=6(x+1) \Rightarrow y=6x+6 \Rightarrow 6x-y=-6
\displaystyle \text{At } (1,0),\ m_{2}=-2
\displaystyle y-0=-2(x-1) \Rightarrow y=-2x+2 \Rightarrow 2x+y=2
\displaystyle \text{At } (2,0),\ m_{3}=3
\displaystyle y-0=3(x-2) \Rightarrow y=3x-6 \Rightarrow 3x-y=6
\\

\displaystyle \textbf{Question 10. }\text{Find the equation of tangent to the curve} \ \
\displaystyle y=\frac{x-7}{x^2-5x+6}\ \text{at the point, where it cuts the X-axis.} \ \text{ \hspace{2.2cm} ISC 2009} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } y=\frac{x-7}{x^{2}-5x+6} \qquad \cdots(i)
\displaystyle \text{On differentiating w.r.t. } x,
\displaystyle \frac{dy}{dx}=\frac{(x^{2}-5x+6)\cdot 1-(x-7)(2x-5)}{(x^{2}-5x+6)^{2}}
\displaystyle \text{[using quotient rule]}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{x^{2}-5x+6-(x-7)(2x-5)}{(x^{2}-5x+6)^{2}}
\displaystyle \text{Also, } (x-7)=y(x^{2}-5x+6)
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{1-(2x-5)y}{x^{2}-5x+6} \qquad \cdots(ii)
\displaystyle \text{Since curve cuts } x\text{-axis, } y=0
\displaystyle \Rightarrow \frac{x-7}{x^{2}-5x+6}=0 \Rightarrow x=7
\displaystyle \therefore \text{Point is } (7,0)
\displaystyle \text{Slope at } (7,0):\ m=\frac{1}{49-35+6}=\frac{1}{20}
\displaystyle \text{Equation of tangent: } y-0=\frac{1}{20}(x-7)
\displaystyle \Rightarrow 20y=x-7 \Rightarrow x-20y=7
\\

\displaystyle \textbf{Question 11. }\text{Find the equations of the normal to the curve} \ \
\displaystyle y=x^3+2x+6,\ \text{which are parallel to line } x+14y+4=0. \ \text{ \hspace{2.2cm} ISC 2008} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } y=x^{3}+2x+6 \qquad \cdots(i)
\displaystyle \text{Line: } x+14y+4=0
\displaystyle \text{On differentiating Eq. (i), } \frac{dy}{dx}=3x^{2}+2
\displaystyle \text{Slope of normal }=-\frac{1}{3x^{2}+2}
\displaystyle \text{Slope of line } x+14y+4=0 \text{ is } -\frac{1}{14}
\displaystyle \text{For parallel lines: } -\frac{1}{3x^{2}+2}=-\frac{1}{14}
\displaystyle \Rightarrow 3x^{2}+2=14 \Rightarrow 3x^{2}=12
\displaystyle \Rightarrow x^{2}=4 \Rightarrow x=\pm 2
\displaystyle \text{When } x=2,\ y=8+4+6=18
\displaystyle \text{When } x=-2,\ y=-8-4+6=-6
\displaystyle \therefore \text{Points are } (2,18),(-2,-6)
\displaystyle \text{Slope of normal }=-\frac{1}{14}
\displaystyle \text{At } (2,18):\ y-18=-\frac{1}{14}(x-2)
\displaystyle \Rightarrow 14y-252=-x+2 \Rightarrow x+14y=254
\displaystyle \text{At } (-2,-6):\ y+6=-\frac{1}{14}(x+2)
\displaystyle \Rightarrow 14y+84=-x-2 \Rightarrow x+14y=-86
\displaystyle \therefore \text{Normals are } x+14y=254 \text{ and } x+14y=-86
\\

\displaystyle \textbf{Question 12. }\text{Find the angle of intersection of the curves } y^2=4ax \ \
\displaystyle \text{and } x^2=4by. \ \text{ \hspace{2.2cm} ISC 2004} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } y^{2}=4ax \qquad \cdots(i)\ \text{and}\ x^{2}=4by \qquad \cdots(ii)
\displaystyle \text{Angle of intersection is angle between tangents at point of intersection}
\displaystyle \text{Substitute } y=\frac{x^{2}}{4b} \text{ in (i):}
\displaystyle \left(\frac{x^{2}}{4b}\right)^{2}=4ax \Rightarrow \frac{x^{4}}{16b^{2}}=4ax
\displaystyle \Rightarrow x^{4}=64ab^{2}x \Rightarrow x(x^{3}-64ab^{2})=0
\displaystyle \Rightarrow x=0 \text{ or } x=4a^{1/3}b^{2/3}
\displaystyle \text{At } x=0,\ y=0
\displaystyle \text{At } x=4a^{1/3}b^{2/3},\ y^{2}=16a^{4/3}b^{2/3} \Rightarrow y=4a^{2/3}b^{1/3}
\displaystyle \text{Points: } (0,0)\ \text{and}\ (4a^{1/3}b^{2/3},4a^{2/3}b^{1/3})
\displaystyle \text{At } (0,0):\ m_{1}=\frac{2a}{y}\to\infty,\ m_{2}=\frac{x}{2b}=0
\displaystyle \Rightarrow \text{Tangents are perpendicular } \Rightarrow \theta=\frac{\pi}{2}
\displaystyle \text{At } (4a^{1/3}b^{2/3},4a^{2/3}b^{1/3}):
\displaystyle m_{1}=\frac{2a}{y}=\frac{2a}{4a^{2/3}b^{1/3}}=\frac{1}{2}\left(\frac{a}{b}\right)^{1/3}
\displaystyle m_{2}=\frac{x}{2b}=\frac{4a^{1/3}b^{2/3}}{2b}=2\left(\frac{a}{b}\right)^{1/3}
\displaystyle \tan\theta=\left|\frac{m_{2}-m_{1}}{1+m_{1}m_{2}}\right|
\displaystyle =\frac{2\left(\frac{a}{b}\right)^{1/3}-\frac{1}{2}\left(\frac{a}{b}\right)^{1/3}}{1+\left(\frac{a}{b}\right)^{2/3}}
\displaystyle =\frac{\frac{3}{2}\left(\frac{a}{b}\right)^{1/3}}{1+\left(\frac{a}{b}\right)^{2/3}}
\displaystyle =\frac{3(ab)^{1/3}}{2(a^{2/3}+b^{2/3})}
\displaystyle \Rightarrow \theta=\tan^{-1}\left(\frac{3(ab)^{1/3}}{2(a^{2/3}+b^{2/3})}\right)
\displaystyle \therefore \text{Angles of intersection are } \frac{\pi}{2} \text{ and } \tan^{-1}\left(\frac{3(ab)^{1/3}}{2(a^{2/3}+b^{2/3})}\right)
\\

\displaystyle \textbf{Question 13. }\text{Find the equation of tangents to the curve} \ \
\displaystyle y=\cos(x+y),\ -2\pi \leq x \leq 2\pi\ \text{that are parallel to the line} \ \text{ \hspace{2.2cm} ISC 2003} \ \
\displaystyle x+2y=0. \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } y=\cos(x+y),\ -2\pi\le x\le 2\pi \qquad \cdots(i)
\displaystyle \text{Line: } x+2y=0 \qquad \cdots(ii)
\displaystyle \text{Slope of line }=-\frac{1}{2}
\displaystyle \text{Let point be } (x_{1},y_{1})
\displaystyle y_{1}=\cos(x_{1}+y_{1})
\displaystyle \text{Differentiate (i): } \frac{dy}{dx}=-\sin(x+y)\left(1+\frac{dy}{dx}\right)
\displaystyle \Rightarrow \frac{dy}{dx}+\sin(x+y)\frac{dy}{dx}=-\sin(x+y)
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{-\sin(x+y)}{1+\sin(x+y)}
\displaystyle \text{At point: } \frac{dy}{dx}=\frac{-\sin(x_{1}+y_{1})}{1+\sin(x_{1}+y_{1})}
\displaystyle -\frac{1}{2}=\frac{-\sin(x_{1}+y_{1})}{1+\sin(x_{1}+y_{1})}
\displaystyle \Rightarrow 1+\sin(x_{1}+y_{1})=2\sin(x_{1}+y_{1})
\displaystyle \Rightarrow \sin(x_{1}+y_{1})=1 \qquad \cdots(iv)
\displaystyle \text{Also, } y_{1}=\cos(x_{1}+y_{1}) \qquad \cdots(iii)
\displaystyle \text{Squaring and adding (iii) and (iv):}
\displaystyle \cos^{2}(x_{1}+y_{1})+\sin^{2}(x_{1}+y_{1})=1+y_{1}^{2}
\displaystyle \Rightarrow 1=1+y_{1}^{2} \Rightarrow y_{1}=0
\displaystyle \text{Putting } y_{1}=0 \text{ in (iii): } \cos x_{1}=0
\displaystyle \Rightarrow x_{1}=\frac{\pi}{2},\frac{3\pi}{2},-\frac{\pi}{2},-\frac{3\pi}{2}
\displaystyle \text{Only } x_{1}=\frac{\pi}{2},-\frac{3\pi}{2} \text{ satisfy (iv)}
\displaystyle \therefore \text{Points are } \left(\frac{\pi}{2},0\right),\left(-\frac{3\pi}{2},0\right)
\displaystyle \text{Equations of tangents:}
\displaystyle y=\frac{1}{2}\left(x-\frac{\pi}{2}\right),\quad y=-\frac{1}{2}\left(x+\frac{3\pi}{2}\right)
\displaystyle \Rightarrow 2x+4y=\pi \text{ and } 2x+4y=-3\pi

\\
\displaystyle \textbf{Question 14. }\text{Find the value of } p \text{ for which the curves} \ \
\displaystyle x^2=9p(9-y)\ \text{and } x^2=p(y+1)\ \text{cut each other at right angles.} \ \text{ \hspace{2.2cm} ISC 2001} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } x^{2}=9p(9-y) \qquad \cdots(i),\ x^{2}=p(y+1) \qquad \cdots(ii)
\displaystyle \text{At intersection: } 9p(9-y)=p(y+1)
\displaystyle \Rightarrow 9(9-y)=y+1 \Rightarrow 81-9y=y+1
\displaystyle \Rightarrow 80=10y \Rightarrow y=8
\displaystyle \text{Substitute in (i): } x^{2}=9p \Rightarrow x=\pm 3\sqrt{p}
\displaystyle \text{Points: } (3\sqrt{p},8),(-3\sqrt{p},8)
\displaystyle \text{From (i): } \frac{x^{2}}{9p}=9-y \Rightarrow y=9-\frac{x^{2}}{9p}
\displaystyle \Rightarrow \frac{dy}{dx}=-\frac{2x}{9p} \qquad \cdots(iii)
\displaystyle \text{From (ii): } \frac{x^{2}}{p}=y+1 \Rightarrow y=\frac{x^{2}}{p}-1
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{2x}{p} \qquad \cdots(iv)
\displaystyle \text{At } (3\sqrt{p},8):\ m_{1}=-\frac{6\sqrt{p}}{9p},\ m_{2}=\frac{6\sqrt{p}}{p}
\displaystyle \text{Since perpendicular: } m_{1}m_{2}=-1
\displaystyle \Rightarrow \left(-\frac{6\sqrt{p}}{9p}\right)\left(\frac{6\sqrt{p}}{p}\right)=-1
\displaystyle \Rightarrow \frac{-36p}{9p^{2}}=-1 \Rightarrow \frac{-4}{p}=-1
\displaystyle \Rightarrow p=4
\displaystyle \text{Similarly at } (-3\sqrt{p},8) \text{ gives same result } p=4
\displaystyle \therefore \text{Required value of } p=4
\\

\displaystyle \textbf{Question 15. }\text{Find the equations of the tangent to the curve} \ \text{ \hspace{3.2cm} ISC 2000} \ \
\displaystyle y=x^2-2x+7\ \text{which is} \ \
\displaystyle \text{(i) parallel to the line } 2x-y+9=0, \ \
\displaystyle \text{(ii) perpendicular to the line } 5y-15x=13. \ \
\displaystyle \text{Answer:}
\displaystyle \text{Given, } y=x^{2}-2x+7 \qquad \cdots(i)
\displaystyle \text{On differentiating: } \frac{dy}{dx}=2x-2
\displaystyle \textbf{(i) } \text{Given, equation of the line is } 2x-y+9=0 \Rightarrow y=2x+9
\displaystyle \text{Slope of the line } m=2
\displaystyle \text{If tangent is parallel to the line, then } \frac{dy}{dx}=2
\displaystyle \text{Given curve: } y=x^{2}-2x+7
\displaystyle \frac{dy}{dx}=2x-2
\displaystyle 2x-2=2 \Rightarrow x=2
\displaystyle \text{At } x=2,\ y=2^{2}-2(2)+7=7
\displaystyle \text{Point }(2,7)
\displaystyle \text{Equation of tangent: } y-7=2(x-2)
\displaystyle \Rightarrow y=2x+3 \Rightarrow 2x-y+3=0
\displaystyle \textbf{Answer: } 2x-y+3=0
\displaystyle \textbf{(ii) } \text{Given line: } 5y-15x=13 \Rightarrow y=3x+\frac{13}{5}
\displaystyle \text{Slope of given line } =3
\displaystyle \text{Slope of perpendicular tangent } =-\frac{1}{3}
\displaystyle \frac{dy}{dx}=2x-2
\displaystyle 2x-2=-\frac{1}{3}
\displaystyle 2x=\frac{5}{3} \Rightarrow x=\frac{5}{6}
\displaystyle y=\left(\frac{5}{6}\right)^{2}-2\left(\frac{5}{6}\right)+7
\displaystyle =\frac{25}{36}-\frac{10}{6}+7=\frac{25-60+252}{36}=\frac{217}{36}
\displaystyle \text{Point } \left(\frac{5}{6},\frac{217}{36}\right)
\displaystyle \text{Equation of tangent: } y-\frac{217}{36}=-\frac{1}{3}\left(x-\frac{5}{6}\right)
\displaystyle 36y-217=-12x+10
\displaystyle \Rightarrow 12x+36y-227=0
\displaystyle \textbf{Answer: } 12x+36y-227=0

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