Question

If the tangent at a point on the ellipse $\frac{x^2}{27}+\frac{y^2}{3}=1$ meets the coordinate axes at A and B, and O is the origin, then the minimum area (in sq. units) of the triangle $OAB$ is:

• A. $\frac{9}{2}$
• B. $9$
• C. $3\sqrt{3}$
• D. $9\sqrt{3}$

Answer 1

Answer: (B)

$9$

$y=mx\pm \sqrt{(a^2{m}^2+b^2)}$
$y=mx\pm \sqrt{(27m^2+3)}$
Intercept on $X$ axis =$\frac{\sqrt{(27m^2+3)}}{m}$
Intercepton $Y$ axis$=\sqrt{(27m^2+3)}$
Area $=\frac{1}{2}\times \frac{\sqrt{(27m^2+3)}}{m}\times \sqrt{(27m^2+3)}$
$=\frac{1}{2}\times \frac{27m^2+3}{m}$
AM $\ge$ GM
$\frac{27m+\frac{3}{m}}{2}\ge \sqrt{27m\times \frac{3}{m}}\ge 9$
Minimum Value $=9$