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 $\frac{\pi }{2}$

Given curve I as

$x^2-y^2=8$

$\Rightarrow \frac{dy}{dx}=\frac{x}{y}$

Slope of tangent at $(-\frac{5}{\sqrt{2}},\frac{3}{\sqrt{2}})=-\frac{5}{3}$

Slope of normal to curve I at this point , $m_1=\frac{3}{5}$

Given curve II as

$9x^2+25y^2=225$

$\Rightarrow \frac{dy}{dx}=-\frac{9x}{25y}$

Slope of tangent at $(-\frac{5}{\sqrt{2}},\frac{3}{\sqrt{2}})=\frac{3}{5}$

Slope of normal to curve II at this point , $m_2=-\frac{5}{3}$

$\theta =ta{n}^{-1}|\frac{m_1-m_2}{1+m_1{m}_2}|$

$\Rightarrow \theta =\frac{\pi }{2}$