Question

The line $y=mx-\frac{\left(a^2-b^2\right)m}{\sqrt{a^2+b^2{m}^2}}$ is normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ for all values of $m$ belonging to

• A. $(0,1)$
• B. $(0,\infty )$
• C. $R$
• D. none of these

Answer 1

Answer: (C)

The equation of the normal to the given ellipse at the point
$P(a\cos \theta ,b\sin \theta )$ is ax $\sec \theta -\mathrm{cosec}\theta =a^2-b^2$. Then,
$y=\left(\frac{a}{b}\tan \theta \right)x-\frac{\left(a^2-b^2\right)}{b}\sin \theta$
Let $\frac{a}{b}\tan \theta =m$
so that
$\sin \theta =\frac{bm}{\sqrt{a^2+b^2{m}^2}}$
Hence, the equation of the normal equation (i) becomes
$y=mx-\frac{\left(a^2-b^2\right)m}{\sqrt{a^2+b^2{m}^2}}$
$\therefore m\in R,\text{ as }m=\frac{a}{b}\tan \theta \in R$