Question

The polar of any point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with respect to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ :

• A. will not intersect the ellipse
• B. will touch at other end of ordinate through the point
• C. will touch on the vertex
• D. none of these

Answer 1

Answer: (B)

will touch at other end of ordinate through the point

Let $P(a\cos \varphi ,b\sin \varphi )$ be any point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ ...(1)
The polar of $P(a\cos \varphi ,b\sin \varphi )$ with respect to $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is
$\frac{x\cos \varphi }{a}-\frac{y\sin \varphi }{b}=1\Rightarrow y=\frac{b}{a}\cot \varphi -b\csc \varphi$
It will touch the ellipse(1), if $(c^2=a^2{m}^2+b^2)$
$b^2{\csc }^2\varphi =a^2\frac{b^2}{a^2}{\cot }^2\varphi +b^2\Rightarrow {\csc }^2\varphi ={\cot }^2\varphi +1$