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
B
18
C
272
D
27

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 $$= 4 \times \dfrac{27}{4} = 27$$ Mathematics

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