Question

If the area of the triangle formed by the tangent at any point P to elipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and axes of ellipse be minimum then point P is $(\frac{a}{\sqrt{2}},\frac{b}{\sqrt{2}}).$

If true enter 1,if false enter 0.

Answer 1

Equation of tangent at $(a\cos \theta ,b\sin \theta )$ to the sllipse is

$\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$

Coordinates of $P$ and $Q$ are

$(\frac{a}{\cos \theta },0)$ and $(0,\frac{b}{\sin \theta })$, respectively

Area of $△OPQ=\frac{1}{2}|\frac{a}{\cos \theta }\times \frac{b}{\sin \theta }|=\frac{ab}{|\sin 2\theta |}\ge ab...[\because |\sin \theta |\le 1]$

$therefore$ Minimum area $=ab$