Tangent at a point of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is drawn which cuts the coordinate axes at $A$ and $B$. The minimum area of the $\mathrm{\Delta}OAB\$is ($O$ being the origin)

Question

Dr. Abubacker · Accepted Answer

(a)
Equation of tangent at $P({acos}{	heta,b\sin	heta)}$ is
$\frac{x}{a}\cos{	heta + \frac{y}{b}\sin{	heta = 1}}$
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Whose point of intersection of axes are
$A\left( \frac{a}{\cos	heta},0 ight)\ and\ B\left( 0,\frac{b}{\sin	heta} ight)$
$	herefore Area\ of\ \mathrm{\Delta}AOB = \frac{1}{2}\ \left| \frac{a}{\cos	heta}.\frac{b}{\sin	heta} ight|$
$\mathrm{\Delta} = \frac{ab}{{|sin}{2	heta}|}$
Now area is minimum when $|\sin{2	heta|\ }$is maximum $ie,|\sin{2	heta| = 1}$
$	herefore\ \ \mathrm{\Delta}_{minimum} = ab$

Q. Tangent at a point of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is drawn which cuts the coordinate axes at $A$ and $B$. The minimum area of the $\mathrm{\Delta}OAB\$is ($O$ being the origin)

Explanation

Reviewed by Dr. Abubacker

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