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VITEEE 2016 Paper

Section: Mathematics

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Question : 83 of 125

Marks: +1, -0

The foci of the ellipse

\frac{x^2}{16}+\frac{y^2}{b^2}

= 1 , and hyperbola

\frac{x^2}{144}-\frac{y^2}{81}

\frac{1}{25}

coincide then value of

b^2

is

1
5
7
9

Solution:

Given ellipse

\frac{x^2}{16}+\frac{y^2}{b^2} = 1

Now

b^2 = a^2(1-e^2)

\Rightarrow b^2 = 16(1-e^2)

,

\Rightarrow \frac{b^2}{16} = 1 - e^2

\Rightarrow e^2 = 1 - \frac{b^2}{16} = \frac{16-b^2}{16}

\Rightarrow e = \frac{\sqrt{16-b^2}}{4}

Foci

= (\neq ae , 0) = (\pm \sqrt{16-b^2},0)

Given hyperbola :

\frac{x^2}{144}-\frac{y^2}{81} = \frac{1}{25}

\Rightarrow \frac{x^2}{\left(\frac{12}{5}\right)^2}-\frac{y^2}{\left(\frac{9}{5}\right)^2} = 1

Now,

b^2 = a^2(e^2-1)

\Rightarrow \left(\frac{9}{5}\right)^2 = \left(\frac{12}{5}\right)^2 (e^2 - 1)

\Rightarrow \left(\frac{9}{12}\right)^2 = e^2 - 1

\Rightarrow e^2 = 1 + \frac{81}{144} = \frac{144+81}{144}

\\

\Rightarrow e = \frac{15}{12} = 5/4

\\ Foci

= (\neq ae, 0) = (\neq 3, 0)

\\ Since foci of the given ellipse and hyperbola coincide, therefore

\sqrt{16-b^2} = 3 \Rightarrow 16 - b^2 = 9

\\

\therefore b^2 = 7

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