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)

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.

N neha

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