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  • Can someone help me with this, - Co-ordinate geometry - JEE Main-3

The foci of the ellipse \frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1  and the hyperbola \frac{x^{2}}{144}-\frac{y^{2}}{81}=\frac{1}{25}     coincide. Then the value of b^{2}  is

  • Option 1)

    5

  • Option 2)

    7

  • Option 3)

    9

  • Option 4)

    1

 

Answers (1)

best_answer

As we learnt in

Eccentricity -

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

- wherein

For the ellipse  

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

 

 and

Coordinates of foci -

\pm ae,o

- wherein

For the ellipse  

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

 

 

\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1, \frac{x^{2}}{\frac{144}{25}}- \frac{y^{2}}{\frac{81}{25}}=1

e_{1}=\sqrt{1-\frac{b^{2}}{16}}\ and \ e_{2}=\sqrt{1+\frac{81}{144}}

So if this foci coincide

a_{1}e_{1}=a_{2}e_{2}

4 \sqrt{16-b^{2}}=\frac{12}{5}\sqrt{1+\frac{81}{144}}

\frac{4\sqrt{16-b^{2}}}{4}=\frac{12}{5}\times \frac{15}{12}

\sqrt{{16-b^{2}}}=3

b^{2}=7 


Option 1)

5

This option is incorrect.

Option 2)

7

This option is correct.

Option 3)

9

This option is incorrect.

Option 4)

1

This option is incorrect.

Posted by

prateek

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