Question

If 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, write the value of b².

Answer 1

Equation of the hyperbola:

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

Simplifying the above equation, we get:

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

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

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

This is the standard form of hyperbola with \[a = \frac{12}{5}\] and  \[b = \frac{9}{5}\].

Eccentricity of the hyperbola, \[e = \frac{\sqrt{\left( \frac{12}{5} \right)^2 + \left( \frac{9}{5} \right)^2}}{\frac{12}{5}}\]

                                                     \[ = \frac{15}{12}\]

The foci of the hyperbola are of the form 

\[\left( ae, 0 \right)\] and \[\left( - ae, 0 \right)\].

So, foci of the hyperbola are  \[\left( 3, 0 \right)\] and  \[\left( - 3, 0 \right)\] .

The foci of the hyperbola coincide with the foci of the ellipse.
For ellipse, \[a = 4\]

Using the relation \[\sqrt{a^2 - b^2} = ae\], we get:

\[\sqrt{4^2 - b^2} = 3\]

\[ \Rightarrow 16 - b^2 = 9\]

\[ \Rightarrow b^2 = 7\]

Therefore, the value of \[b^2\] is 7.