Question

If the tangent to the ellipse $x^2+4y^2=16$ at a point $(4\cos \theta ,2\sin \theta )$ passes through the focus of the parabola $x^2=8(y-6)$, then

• A. $\sin \theta =\frac{1}{4}$
• B. $\sin \theta =\frac{2}{3}$
• C. $\tan \theta =\frac{1}{4}$
• D. $\cot \theta =\frac{1}{3}$

Answer 1

The correct option is A $\sin \theta =\frac{1}{4}$

Given equation is $x^2+4y^2=16$

Therefore, wee have $\frac{x^2}{16}+\frac{y^2}{4}=1$
The tangent at $P\left(\theta \right)$ is $\frac{x}{4}\cos \theta +\frac{y}{2}\sin \theta =1$
The focus of the parabola $x^2=8(y-6)$ is $(0,8)$.
The focus lies on the tangent. Hence,
$\therefore 0+\frac{8}{2}\sin \theta =1\Rightarrow \sin \theta =\frac{1}{4}$
Hence, option A is correct.