Question

The angle between the curves $y^2=4x+4$ and $y^2=36(9-x)$ is?

• A. ${30}^o$
• B. ${45}^o$
• C. ${60}^o$
• D. ${90}^o$(orthogonal curves)

Answer 1

Answer: (D)

${90}^o$(orthogonal curves)

$y^2=4(x+1)$

$y^2=36(9-x)$

$⟹4(x+1)=36(9-x)$

$⟹x+1=9(9-x)$

$⟹x+1=81-9x$

$⟹10x=80$

$⟹x=8$

$y^2=4(x+1)$

$⟹y^2=4(8+1)$

$⟹y^2=36$

$⟹y=\pm 6$

$\therefore pointsofintersectionare(8,6)\&(8,-6)$

$y^2=4x+4$

$differentiating$

$⟹2y\frac{dy}{dx}=4$

$⟹y\frac{dy}{dx}=2$

$⟹\frac{dy}{dx}=\frac{2}{y}$

$\therefore At(8,6),\frac{dy}{dx}=\tan \theta 1=\frac{2}{6}=\frac{1}{3}$

$At(8,-6),\frac{dy}{dx}=\tan \theta 1^1=\frac{-2}{6}=\frac{-1}{3}$

$y^2=324-36x$

$⟹2y\frac{dy}{dx}=-36$

$⟹\frac{dy}{dx}=\frac{-18}{y}$

$At(8,6),\tan \theta 2=\frac{-18}{6}=-3$

$At(8,-6),\tan \theta 2^1=\frac{-18}{-6}=3$

$\therefore Angleofintersectionat(8,6)$

$\tan \theta =\frac{\tan \theta 1-\tan \theta 2}{1+\tan \theta 1.\tan \theta 2}$

$=\frac{\frac{1}{3}-(-3)}{1+\left(\frac{1}{3}\right(-3\left)\right)}$

$=\frac{\frac{1}{3}+3}{1-1}$

$\therefore \theta ={90}^{\circ }$

$Angleofintersectionat(8,-6)$

$\tan \theta =\frac{\frac{-1}{3}-3}{1+\left(\frac{-1}{3}\right(3\left)\right)}$

$⟹\theta ={90}^{\circ }$

$\therefore$ Two curves are orthogonal

$\therefore D){90}^{\circ }$