Question

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

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

Answer 1

The correct option is A $ab$ sq. units

Any tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at $P(a\cos \theta ,b\sin \theta )$ is given by
$\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1$


It meets coordinate axes at $A(a\sec \theta ,0)$ and $B(0,b\text{cosec}\theta )$
$\therefore$ Area of $△OAB=\frac{1}{2}\times |a\sec \theta \times b\text{cosec}\theta |$
$\Rightarrow △=\left|\frac{ab}{\sin 2\theta }\right|$

For area to be minimum $\sin 2\theta$ should be maximum and we know that maximum value of $\sin 2\theta$ is $1.$
$\therefore {△}_{max}=ab$ Sq.units