NIMCET 2014 Question Paper with solutions

© examsiri.com

Question : 28 of 120

Marks: +1, -0

If the foci of the ellipse

b^2x^2

+

16y^2

-

16b^2

and the hyperbola

81x^2

- 144

y^2

=

{81×144}/{25}

coincide, then the value of b is

1
$√5$
$√7$
3

Solution:

Since, foci coincide
∴

16-b^2

=

{144}/{25}

+

{81}/{25}

⇒

16-b^2

=

{225}/{25}

⇒

b^2

= 16 -

{225}/{25}

=

{400-225}/{25}

=

{175}/{25}

= 7
∴ b =

√7

© examsiri.com

Go to Question:

Prev Question

Next Question