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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Coordinate Geometry

Find the maximum area of $\Delta PSS'$ for the ellipse $\large\frac{x^2}{36}+\frac{y^2}{16}$$=1$ where S,S' are foci & P is any point on ellipse?

$\begin{array}{1 1}(a)\;8\sqrt 6\\(b)\;8\sqrt 2\\(c)\;8\sqrt 5\\(d)\;\text{None of these}\end{array}$

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1 Answer

$a^2=36$
$b^2=16$
$\large\frac{b^2}{a^2}$$=1-e^2$
$\large\frac{16}{36}$$-1=-e^2$
$\large\frac{20}{36}$$=-e^2$
$e=\large\frac{2\sqrt 5}{6}$
$e=\large\frac{\sqrt 5}{3}$
For maximum area of $\Delta PSS'$ P should be one of the vertex of minor axis hence
Area of $\Delta PSS'=\large\frac{1}{2}$$2\times ae\times b=aeb$
$\Rightarrow 6\times 4 \times \large\frac{\sqrt 5}{3}$
$\Rightarrow 8\sqrt 5$
Hence (c) is the correct answer.
answered Feb 7, 2014 by sreemathi.v
edited Sep 23, 2014 by sharmaaparna1
 

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