Answer:
Option B
Explanation:
Given equation of curves are
$x^{2}=8y$ .......(i)
and xy=8 ..........(ii)
On substituting the value of y from Eq.(i) in Eq.(ii) ,
we get
$x\left(\frac{x^{2}}{8}\right)=8$
$\Rightarrow$ $x^{3}=64$
$\Rightarrow$ x=4
On substituting x=4 in Eq.(ii) , we get y=2
Thus , the point of intersection of given curves is (4,2)
Now, let $m_{1}$ be the slope of tangent to the curve (i) at point (4,2) and $m_{2}$ be the slope of tangent to the curve (ii)
at point (4,2)
Then, $m_{1}=\frac{4}{4}=1$
$\left[ \because x^{2}=8y \Rightarrow 2x=8\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{x}{4}\right]$
and $m_{2}= \frac{-2}{4}=-\frac{1}{2}$
$\left[ \because xy=8 \Rightarrow x\frac{dy}{dx}+y=0\Rightarrow \frac{dy}{dx}=\frac{-y}{4}\right]$
Now, the angle $\theta$ between the curves is given by
$\tan \theta= \left(\frac{m_{2}-m_{1}}{1+m_{1}m_{2}}\right)$
= $ \left(\frac{\frac{-1}{2}-1}{1-\frac{1}{2}}\right)=\left(\frac{\frac{-3}{2}}{\frac{1}{2}}\right)$
$\Rightarrow$ $\tan \theta=-3$
$\Rightarrow$= $\theta= tan^{-1}(-3)$
