Question

If the minimum area of the triangle formed by a tangent to the ellipse $\frac{x^2}{b^2}+\frac{y^2}{4a^2}=1$ and the coordinate axis is $kab$, then $k$ is equal to

Answer 1

Any point on ellipse $\frac{x^2}{b^2}+\frac{y^2}{4a^2}=1$ can be taken as $P(b\cos \theta ,2a\sin \theta ).$
Tangent at $P:\frac{x}{b}\cos \theta +\frac{y}{2a}\sin \theta =1$
At $x=0\Rightarrow y=\frac{2a}{\sin \theta }$
At $y=0\Rightarrow x=\frac{b}{\cos \theta }$
Area of triangle
$=\frac{1}{2}|\frac{2a}{\sin \theta }\times \frac{b}{\cos \theta }|=\left|\frac{2ab}{\sin 2\theta }\right|$
Minimum area $=2ab$ when $\sin 2\theta =\pm 1$
$\therefore k=2$.