  Question

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

A
274  B
18  C
272  D
27  Solution

The correct option is D $$27$$Given equation of ellipse is $$\dfrac{{x}^{2}}{9}+\dfrac{{y}^{2}}{5}=1$$       .....$$(1)$$$$\therefore {a}^{2}=9,{b}^{2}=5$$$$\Rightarrow a=3,b=\sqrt{5}$$Now, $$e=\sqrt{1-\dfrac{{b}^{2}}{{a}^{2}}}=\sqrt{1-\dfrac{5}{9}}=\sqrt{\dfrac{9-5}{9}}=\dfrac{2}{3}$$Foci$$=\left(\pm ae,0\right)=\left(\pm 2,0\right)$$and $$\dfrac{{b}^{2}}{a}=\dfrac{5}{3}$$$$\therefore$$ Extremities of one of latus rectum are $$\left(2,\dfrac{5}{3}\right)$$ and $$\left(2,\dfrac{-5}{3}\right)$$$$\therefore$$ Equation of tangent at $$\left(2,\dfrac{5}{3}\right)$$ is $$\dfrac{x\left(2\right)}{9}+\dfrac{y\left(\dfrac{5}{3}\right)}{5}=1$$or $$2x+3y=9$$         ........$$(2)$$Eqn$$(2)$$ intersects $$X$$ and $$Y-$$ axes at $$\left(\dfrac{9}{2},0\right)$$ and $$\left(0,3\right)$$ respectively.$$\therefore$$ Area of quadrilateral$$=4\times$$Area of $$\triangle{POQ}$$$$=4\times\dfrac{1}{2}\times \dfrac{9}{2}\times 3=27$$sq.units.Maths

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