Question

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 b² is
1. 1
2. 5
3. 7
4. 9

Answer 1

Correct Answer - Option 3 : 7

Concept:

Eccentricity for the ellipse \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\)  is given by \(e = \sqrt {1 - \frac{{{b^2}}}{{{a^2}}}} \)   and foci = (±ae, 0)

Eccentricity for the hyperbola \(\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\) is given by \(e = \sqrt {1 + \frac{{{b^2}}}{{{a^2}}}} \)   and foci = (± ae, 0)

Calculation:

\(\frac {x^2}{16} + \frac {y^2}{b^2} = 1\)   eccentricity of the ellipse is \(e = \sqrt {1 - \frac{{{b^2}}}{{16}}} \)  and x coordinate foci = \(4\sqrt {1 - \frac{{{b^2}}}{{16}}} = \sqrt {16 - {b^2}} \)

Given hyperbola \(\frac {x^2}{144} - \frac {y^2}{81} = \frac 1 {25}\)  can be written as \(\frac{{{x^2}}}{{{(12/5)^2}}} - \frac{{{y^2}}}{{{(9/5)^2}}} = 1\)  ,

By comparing the given hyperbola with \(\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\), we get a = 12/5,   b = 9/5 

Eccentricity of the hyperbola

\(e = \sqrt {1 + \frac{{{b^2}}}{{{a^2}}}} ⇒ \sqrt {1 + \frac{{{{(9/5)}^2}}}{{{{(12/5)}^2}}}} ⇒ \sqrt {1 + \frac{{81}}{{144}}} ⇒ \frac{{15}}{{12}}\)

x coordinate foci of hyperbola = ae = \(\frac{{12}}{5} \times \frac{{15}}{{12}} = 3\)

According to question foci of ellipse and hyperbola coincide

So x coordinate of foci of ellipse equal to the x coordinate of foci of hyperbola

\(⇒ \sqrt {16 - {b^2}} = 3\)

⇒ 16 - b² = 9

⇒ b² = 7