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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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The minimum area of triangle formed by the tangents to the ellipse $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ coordinate axes is :

$\begin{array}{1 1}(A)\;ab\;sq.units \\(B)\;\frac{(a+b)^2}{2} \;sq.units \\(C)\;\frac{a^2+b^2}{2}\;sq.units \\(D)\;\frac{a^2+ab+b^2}{3} \end{array}$

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1 Answer

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Equation of tangent to given ellipse is
$y= mx +\sqrt {a^2m^2+b^2}$
It meets axes at $\bigg(-\large\frac{\sqrt {a^2m^2 +b^2}}{m},$$0 \bigg)$ and $ (0, \sqrt {a^2m^2+b^2})$
$\therefore $ Area of triangle
$\qquad= \large\frac{1}{2} \bigg| \large\frac{a^2m^2 +b^2}{m} \bigg|$
$\qquad= \large\frac{1}{2} $$ \bigg| a^2m +\large\frac{b^2}{m} \bigg| $$ \geq ab$
Hence A is the correct answer.
answered Apr 10, 2014 by meena.p
 

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