Question

The equation of normal to the curve $x=a\sin 2\theta ,y=a\cos \theta$ at $\theta =\frac{\pi }{6}$ is

• A. $y=2x+\frac{\sqrt{3}}{2}a$
• B. $2y=x-\frac{\sqrt{3}}{2}a$
• C. $y=2x-\frac{\sqrt{3}}{2}a$
• D. $y=2x-\frac{3}{2}a$

Answer 1

The correct option is C $y=2x-\frac{\sqrt{3}}{2}a$

$x=a\sin 2\theta ,y=a\cos \theta$
Differentiating w.r.t. $\theta$, we get,
$\frac{dx}{d\theta }=2a\cos 2\theta$
$\frac{dy}{d\theta }=-a\sin \theta$
Slope of the tangent at $\theta =\frac{\pi }{6}$ is
$m=\frac{dy}{dx}{|}_{\theta =\frac{\pi }{6}}=\frac{-\frac{a}{2}}{a}=-\frac{1}{2}$
Slope of the normal
$=\frac{-1}{m}=2$
Point of contact
$x=a\frac{\sqrt{3}}{2}$ and $y=a\frac{\sqrt{3}}{2}$
Equation of the normal
$\frac{y-\frac{\sqrt{3}}{2}a}{x-\frac{\sqrt{3}}{2}a}=2\Rightarrow y=2x-\frac{\sqrt{3}}{2}a$