Question

The angle of intersection of the normals at the point $(-\frac{5}{\sqrt{2}},\frac{3}{\sqrt{2}})$ of the curves $x^2-y^2=8$ and $9x^2+25y^2=225$ is

• A. $0$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is B $\frac{\pi }{2}$

$x^2-y^2=8$
$\Rightarrow \frac{dy}{dx}=\frac{x}{y}$
$\Rightarrow -\frac{1}{dy/dx}=\frac{y}{x}$

At the point $(-\frac{5}{\sqrt{2}},\frac{3}{\sqrt{2}}),$
$-\frac{1}{dy/dx}=\frac{-3/\sqrt{2}}{-5/\sqrt{2}}=\frac{3}{5}$

Also, $9x^2+25y^2=225$
$\text{or}18x+50y\frac{dy}{dx}=0$
$\text{or}\frac{dy}{dx}=-\frac{9x}{25y}\text{or}-\frac{dx}{dy}=\frac{25y}{9x}$

At the point $(-\frac{5}{\sqrt{2}},\frac{3}{\sqrt{2}}),$
$-\frac{dx}{dy}=\frac{25\times 3/\sqrt{2}}{9(-5/\sqrt{2})}=-\frac{15}{9}=-\frac{5}{3}$

Since, the product of the slopes is $-1$, the normals cut orthogonally, i.e., the required angle is equal to $\frac{\pi }{2}.$