JEE Main 2015 — Ellipse Question with Solution

The end point of latus rectum of ellipse

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ in first quadrant is

$$\left(ae,\frac{b^2}{a}\right)$$ and the tangent at this point

intersects $$x$$-axis at $$\left(\frac{a}{e},0\right)$$ and

$$y$$-axis at $$(0,a).$$

The given ellipse is $$\frac{x{}^2}{9}+\frac{y^2}{5}=1$$

Then $$a^2=9,b^2=5$$

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

$$\therefore$$ end point of latus rectum in first quadrant is

$$L\left(2,5/3\right)$$

Equation of tangent at $$L$$ is $$\frac{2x}{9}+\frac{y}{3}-1$$

It meets $$x$$-axis at $$A(9/2,0)$$

and $$y$$-axis at $$B(0,3)$$

$$\therefore$$ Area of $$\Delta OAB=\frac{1}{2}\times \frac{9}{2}\times 3=\frac{27}{4}$$


By symmetry area of quadrilateral

$$=4\times \left(Area\Delta OAB\right)=4\times \frac{27}{4}=27$$ sq. units.