Question

If the normal at the point $P\left(\theta \right)$ to the ellipse $\frac{x^2}{3}+\frac{y^2}{2}=1$ intersects it again at the point $Q\left(2\theta \right),$ then find cos$\theta .$

• A. $0$
• B. $1$
• C. $-1$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is C $-1$

Given, equation of ellipse $\frac{x^2}{3}+\frac{y^2}{2}=1$

Equation of normal in parametric form: $\frac{\sqrt{3}x}{\cos \theta }-\frac{\sqrt{2}y}{\sin \theta }=3-2=1$

Since, it passes through $Q$

Therefore, $\frac{3\cos 2\theta }{\cos \theta }-\frac{2\sin 2\theta }{\sin \theta }=1$

$\Rightarrow \frac{3(2{\cos }^2\theta -1)}{\cos \theta }-4\cos \theta =1$

$\Rightarrow 2{\cos }^2\theta -\cos \theta -3=0$

$\Rightarrow \cos \theta =-1$