Question

The value of $\lambda$, for which the line $2x-\frac{8}{3}\lambda y=-3$ is a normal to the ellipse $x^2+\frac{y^2}{4}=1$, is:

• A. $\frac{\sqrt{3}}{2}$
• B. $\frac{1}{2}$
• C. $-\frac{\sqrt{3}}{2}$
• D. $\frac{3}{8}$

Answer 1

Answer: (A), (C)

$\frac{\sqrt{3}}{2}$
$-\frac{\sqrt{3}}{2}$

For the given ellipse $a=1$, $b=2$.

The slope of tangent at point with eccentric angle $\theta$ is $\frac{-b}{a\tan \theta }$.

Hence, the slope of the normal at eccentric angle $\theta$ is $\frac{a\tan \theta }{b}=\frac{\tan \theta }{2}$.

Hence, the equation of the normal is $y-2\sin \theta =\frac{\tan \theta }{2}(x-\cos \theta )$.

$\Rightarrow \frac{\tan \theta }{2}x-y+\frac{3\sin \theta }{2}=0$....(1)

The equation of the normal is given as
$2x-\frac{8}{3}\lambda y+3=0$....(2)

Comparing the coefficients of (1) and (2) we get,

$\frac{\tan \theta }{4}=\frac{3}{8\lambda }=\frac{\sin \theta }{2}$

On solving for $\theta$, we get $\cos \theta =\frac{1}{2}\Rightarrow \sin \theta =\pm \frac{\sqrt{3}}{2}$

Now, using the value of $\sin \theta$ we get $\lambda =\pm \frac{\sqrt{3}}{2}$