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Question

The area (in sq. units) of the quadrilateral formed by the tangent at the end points of the latera  recta to the ellipse $$\displaystyle \dfrac {x^2}{9}+\dfrac {y^2}{5}=1$$, is


A
274
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B
18
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C
272
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D
27
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Solution

The correct option is C $$27$$
For the given ellipse 

$$e = \sqrt { 1 - \dfrac{5}{9} }  = \dfrac{2}{3} $$

The endpoint of one of the latus rectum will be $$ \left (2,\dfrac53 \right) $$

The tangents drawn at the end of the latera recta to the ellipse will intersect at the $$x$$ and the $$y$$ axis because of symmetry. The area of the quadrilateral thus formed will be $$4$$ times the area of the triangle formed by the tangent and the coordinate axis. 
The equation of the tangent at $$\left(2,\dfrac{5}{3} \right) $$ can be written as :
$$ \dfrac{2x}{9} + \dfrac{5y}{15} = 1 $$

X intercept  $$ =\dfrac{9}{2} $$

Y intercept  $$ = 3 $$

Area of the triangle $$ = \dfrac12 \times \dfrac92 \times 3 =\dfrac{27}{4} $$

Hence, area of the quadrilateral $$ = 4 \times \dfrac{27}{4} = 27 $$ 

271850_299511_ans.png

Mathematics

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