Question

The minimum area of the triangle formed by any tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with the coordinate axes is

• A. $a^2+b^2$
• B. $\frac{(a+b{)}^2}{2}$
• C. $ab$
• D. $\frac{(a-b{)}^2}{2}$

Answer 1

The correct option is C $ab$

Equation of tangent at $(acos\theta ,bsin\theta )$ to the ellipse 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 $\Delta OPQ=\frac{1}{2}|\frac{a}{cos\theta }\times \frac{b}{sin\theta }|=\frac{ab}{|sin2\theta |}\ge ab$ .......... [$\because |sin\theta |\le 1$]

$\therefore$ Minimum area $=ab$

Hence, option C is correct.