Question

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

• A. $absq.units$
• B. $\frac{(a+b{)}^2}{2}sq.units$
• C. $\frac{a^2+b^2}{2}sq.units$
• D. $\frac{a^2+ab+b^2}{3}$

Answer 1

The correct option is A

$absq.units$

Equation of tangent to given ellipse is
$y=mx+\sqrt{a^2{m}^2+b^2}$
It meets axes at $(-\frac{\sqrt{a^2{m}^2+b^2}}{m},0)\text{and}(0,\sqrt{a^2{m}^2+b^2})$
$\therefore$ Area of triangle
$=\frac{1}{2}\left|\frac{a^2{m}^2+b^2}{m}\right|$
$=\frac{1}{2}|a^{2}m+\frac{b^2}{m}|\ge ab$
(Using the inequality arithmetic mean greater than or equal to geometric mean)
Hence A is the correct answer.