Question
The area of the quadrilateral formed by the tangents at the end points of latus rectum to the ellipse $$\frac{{{x^2}}}{9} + \frac{{{y^2}}}{5} = 1,$$ is-
A.
$$\frac{{27}}{4}\,\,{\text{sq}}{\text{.}}\,{\text{units}}$$
B.
$$9\,\,{\text{sq}}{\text{.}}\,{\text{units}}$$
C.
$$\frac{{27}}{2}\,\,{\text{sq}}{\text{.}}\,{\text{units}}$$
D.
$$27\,\,{\text{sq}}{\text{.}}\,{\text{units}}$$
Answer :
$$27\,\,{\text{sq}}{\text{.}}\,{\text{units}}$$
Solution :
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 $$L\left( {2,\,\frac{5}{3}} \right)$$
Equation of tangent at $$L$$ is $$\frac{{2x}}{9} + \frac{y}{3} = 1$$
It meets $$x$$-axis at $$A\left( {\frac{9}{2},\,0} \right)$$ and $$y$$-axis at $$B\left( {0,\,3} \right)$$
$$\therefore $$ Area of $$\Delta OAB = \frac{1}{2} \times \frac{9}{2} \times 3 = \frac{{27}}{4}$$
By symmetry area of quadrilateral
$$ = 4 \times \left( {{\text{Area of }}\Delta OAB} \right) = 4 \times \frac{{27}}{4} = 27{\text{ sq}}{\text{. units}}$$
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 $$L\left( {2,\,\frac{5}{3}} \right)$$
Equation of tangent at $$L$$ is $$\frac{{2x}}{9} + \frac{y}{3} = 1$$
It meets $$x$$-axis at $$A\left( {\frac{9}{2},\,0} \right)$$ and $$y$$-axis at $$B\left( {0,\,3} \right)$$
$$\therefore $$ Area of $$\Delta OAB = \frac{1}{2} \times \frac{9}{2} \times 3 = \frac{{27}}{4}$$
By symmetry area of quadrilateral
$$ = 4 \times \left( {{\text{Area of }}\Delta OAB} \right) = 4 \times \frac{{27}}{4} = 27{\text{ sq}}{\text{. units}}$$