The value of , for which the line 2x - 8 3 y = - 3 is a normal to… - Vidyadip

MCQ

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

A
$\frac{{\sqrt 3 }}{2}$
B
$\frac{1}{2}$
C
$ - \frac{{\sqrt 3 }}{2}$
✓
$\frac{3}{8}$

✓

Answer

Correct option: D.

$\frac{3}{8}$

d
(d) We know that the equation of the normal at point $(a\cos \theta ,\,b\sin \theta )$ on the curve ${x^2} + \frac{{{y^2}}}{4} = 1$ is given by

$ax\sin \theta - by{\rm{cosec }}\theta = {a^2} - {b^2}$.....$(i)$

Comparing equation $(i)$ with $2x - \frac{8}{3}\lambda y = - 3$. We get,

$a\sin \theta = 2$, $b{\rm{ cosec}}\theta = \frac{8}{3}\lambda $ or $ab = \frac{{16}}{3}\lambda $.....$(ii)$

$\because \,a = 1,\,b = 2$; $2 = \frac{{16}}{3}\lambda $ or $\lambda = 3/8$

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