Question
The area of the quadrilateral formed by tangents at the end points of latus recta of the ellipse $$\frac{{{x^2}}}{9} + \frac{{{y^2}}}{5} = 1$$ is :
A.
$$\frac{{27}}{4}{\text{ uni}}{{\text{t}}^2}$$
B.
$$9\,{\text{uni}}{{\text{t}}^2}$$
C.
$$\frac{{27}}{2}{\text{ uni}}{{\text{t}}^2}$$
D.
$$27\,{\text{uni}}{{\text{t}}^2}$$
Answer :
$$27\,{\text{uni}}{{\text{t}}^2}$$
Solution :
$$\eqalign{ & {\text{Here, }}{a^2} = 9,\,\,{b^2} = 5 \cr & {\text{So, latus rectum}} = \frac{{2{b^2}}}{a} = \frac{{2 \times 5}}{3} \cr & \therefore \,P = \left( {\alpha ,\,\frac{5}{3}} \right){\text{.}}\,{\text{It is on the ellipse}}{\text{.}} \cr & {\text{So, }}\frac{{{\alpha ^2}}}{9} + \frac{{{{\left( {\frac{5}{3}} \right)}^2}}}{5} = 1{\text{ or }}\alpha = \pm 2.\,\,\,\,{\text{So, }}P = \left( {2,\,\frac{5}{3}} \right) \cr & {\text{The tangent at }}P{\text{ is }}\frac{{2x}}{9} + \frac{{\frac{5}{3}y}}{5} = 1{\text{ or }}\frac{x}{{\frac{9}{2}}} + \frac{y}{3} = 1 \cr & \therefore \,\,{\text{area}} = 4 \times {\text{ar}}\left( {\Delta ROQ} \right) = 4 \times \left( {\frac{1}{2} \times \frac{9}{2} \times 3} \right) = 27 \cr} $$
$$\eqalign{ & {\text{Here, }}{a^2} = 9,\,\,{b^2} = 5 \cr & {\text{So, latus rectum}} = \frac{{2{b^2}}}{a} = \frac{{2 \times 5}}{3} \cr & \therefore \,P = \left( {\alpha ,\,\frac{5}{3}} \right){\text{.}}\,{\text{It is on the ellipse}}{\text{.}} \cr & {\text{So, }}\frac{{{\alpha ^2}}}{9} + \frac{{{{\left( {\frac{5}{3}} \right)}^2}}}{5} = 1{\text{ or }}\alpha = \pm 2.\,\,\,\,{\text{So, }}P = \left( {2,\,\frac{5}{3}} \right) \cr & {\text{The tangent at }}P{\text{ is }}\frac{{2x}}{9} + \frac{{\frac{5}{3}y}}{5} = 1{\text{ or }}\frac{x}{{\frac{9}{2}}} + \frac{y}{3} = 1 \cr & \therefore \,\,{\text{area}} = 4 \times {\text{ar}}\left( {\Delta ROQ} \right) = 4 \times \left( {\frac{1}{2} \times \frac{9}{2} \times 3} \right) = 27 \cr} $$