CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
18
loader
C
272
loader
D
27
loader

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image