Question

The angle between the curves $x^2=8y$ and $xy=8$ is

• A. ${\tan }^{-1}\left(\frac{-1}{3}\right)$
• B. ${\tan }^{-1}(-3)$
• C. ${\tan }^{-1}(-\sqrt{3})$
• D. ${\tan }^{-1}\left(\frac{-1}{\sqrt{3}}\right)$

Answer 1

The correct option is B ${\tan }^{-1}(-3)$

Angle between the curves is defined as the angle between tangents to both the curves at their point of intersection.

Solving both the curves, $xy=8=>x=\frac{8}{y}$

Putting this in $x^2=8y=>\frac{64}{y^2}=8y=>y=2$

=> $x=\frac{8}{y}=\frac{8}{2}=4$

Point of intersection of both the curves is $(4,2)$

Slope of $xy=8$ is given by $m_1=\frac{dy}{dx}=\frac{-8}{x^2}{|}_{x=4}=\frac{-1}{2}$

Slope of $x^2=8y$ is given by $m_2=\frac{dy}{dx}=\frac{x}{4}{|}_{x=4}=1$

Angle $\theta$ between two tangents can be expressed as

$tan\theta =\frac{m_1-m_2}{1+m_1.m_2}=\frac{-3/2}{1/2}=-3$

=> $\theta =ta{n}^{-1}(-3)$