Question

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

• A. ${\sin }^{-1}\left(\frac{1}{4}\right)$
• B. ${\cos }^{-1}\left(\frac{1}{3}\right)$
• C. ${\tan }^{-1}\left(\frac{1}{4}\right)$
• D. ${\cot }^{-1}\left(\frac{1}{3}\right)$

Answer 1

The correct option is A ${\sin }^{-1}\left(\frac{1}{4}\right)$

$\frac{x^2}{16}+\frac{y^2}{4}=1$

The tangent at $\theta$ is
$\frac{x}{4}\cos \theta +\frac{y}{2}\sin \theta =1$
The focus of the parabola $x^2=8(y-6)$, is $(0,8)$.
$\therefore 0+\frac{8}{2}\sin \theta =1\Rightarrow \sin \theta =\frac{1}{4}$