The area (in sq.

Question

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

Accepted Answer

The end point of latus rectum of ellipse

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

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

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

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

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

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

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

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

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

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

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

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

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

By symmetry area of quadrilateral

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

Solutions · 1