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The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse

$\frac{x^{2}}{9}+\frac{y^{2}}{5}=1,\, is:$

Option 1)
$\frac{27}{4}$

Option 2)
18

Option 3)
$\frac{27}{2}$

Option 4)
27

Answers (2)

best_answer

As we have discussed in

Length of latus rectum of ellipse -

$\frac{2b^{2}}{a}$

- wherein

For the ellipse

$\frac{x^{2}}{a^{2}}+ \frac {y^{2}}{b^{2}}= 1$

Coordinates of foci -

$\pm ae,o$

- wherein

For the ellipse

$\frac{x^{2}}{a^{2}}+ \frac {y^{2}}{b^{2}}= 1$

Eccentricity -

$e= \sqrt{1-\frac{b^{2}}{a^{2}}}$

- wherein

For the ellipse

$\frac{x^{2}}{a^{2}}+ \frac {y^{2}}{b^{2}}= 1$

$\frac{x^{2}}{9}+ \frac{y^{2}}{5}= 1$

Equation of tangent is

$\frac{xx_{1}}{9}+ \frac{yy_{1}}{5}=1$

a² = 9, b² = 5

$e=\sqrt{1-\frac{b^{2}}{a^{2}}}= \frac{2}{3}$

Hence, endpoints of LRs = $\left ( \pm 2,\:\pm \frac{5}{3} \right )\:\left [ \pm ae,\:\pm \frac{b^{2}}{a} \right ]$

Thus, equation of one of tangent is

$\frac{2x}{9}+ \frac{y}{3}= 1$

If y = 0 $\Rightarrow x=\frac{9}{2}$

If x = 0 $\Rightarrow y=3$

Hence, area of $\bigtriangleup AOB=\frac{1}{2}\times \frac{9}{2}\times 3=\frac{27}{4}$

Area of quad $ABCD = 4\times \frac{27}{4}= 27\:sq.\:units$

Option 1)

$\frac{27}{4}$

This option is incorrect.

Option 2)

18

This option is incorrect.

Option 3)

$\frac{27}{2}$

This option is incorrect.

Option 4)

27

This option is correct.

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