Question

The line $y=m x-\frac{\left(a^{2}-b^{2}\right) m}{\sqrt{a^{2}+b^{2} m^{2}}}$ 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 $\sec \theta-\operatorname{cosec} \theta= a ^{2}- b ^{2}$. 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
$\sin \theta=\frac{b m}{\sqrt{a^{2}+b^{2} m^{2}}}$
Hence, the equation of the normal equation (i) becomes
$y=m x-\frac{\left(a^{2}-b^{2}\right) m}{\sqrt{a^{2}+b^{2} m^{2}}}$
$\therefore m \in R , \text { as } m =\frac{ a }{ b } \tan \theta \in R$