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The foci of the ellipse \(\rm \frac {x^2}{16} + \frac {y^2}{b^2} = 1\) and the hyperbola \(\rm \frac {x^2}{144} - \frac {y^2}{81} = \frac {1}{25}\) coincide. Then the value of b2 is
1. 5
2. 7
3. 9
4. 1

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Correct Answer - Option 2 : 7

Concept:

The eccentricity of the curve \(\rm \frac {x^2}{a^2} + \frac {y^2}{b^2} = 1\) is \(\rm e^2 = 1-\left({b^2\over a^2}\right)\)

The eccentricity of the curve \(\rm \frac {x^2}{a^2} - \frac {y^2}{b^2} = 1\) is \(\rm e^2 = 1+\left({b^2\over a^2}\right)\)

Foci of Hyperbola and Ellipse are (ae, 0) and (4e, 0)

 

Calculation:

For given Hyperbola \(\rm \frac {x^2}{144} - \frac {y^2}{81} = \frac {1}{25}\)

⇒ \(\rm \frac {25}{144}x^2 - \frac {25}{81}y^2 = 1\)

∴ \(\rm {a_h}^2 = \frac{144}{25}\) and \(\rm {b_h}^2 = \frac{81}{25}\)

Eccentricity of hyperbola 

\(\rm {e_h}^2 = 1+\left({{b_h}^2\over {a_h}^2}\right)\)

⇒ \(\rm {e_h}^2 = 1+\left({81\over 144}\right)\)

⇒ \(\rm {e_h}^2 = \frac{225}{144}⇒\boldsymbol{\rm e_h=\frac{15}{12}}\)

Focus of hyperbola Fh = (aheh, 0), Where eh is the eccentricity of the hyperbola.

⇒ Fh = \(\rm \left(\left(\frac{12}{5}\times\frac{15}{12}\right), 0\right)\) 

⇒ Fh = (3, 0)

 

For given Ellipse \(\rm\frac {x^2}{16} + \frac {y^2}{b^2} =1\)

∴ \(\rm {a_e}^2 = 16\) and \(\rm {b_e}^2 = b^2\)

Focus of ellipse Fe = (ae, 0) = (4ee, 0), Where ee is eccentricity of the ellipse.

Given Focus of ellipse Fe = Fh 

⇒ (4ee, 0) = (3, 0)

⇒ 4ee = 3 ⇒ ee = \(\rm \boldsymbol {3\over4}\)

Also Eccentricity of an ellipse

\(\rm {e_e}^2 = 1-\left({{b_e}^2\over {a_e}^2}\right)\)

⇒ \(\rm \left({3\over4}\right)^2 = 1-\left({{b}^2\over {4}^2}\right)\)

⇒ \(\rm 1-{9\over16} = {{b}^2\over16}\)

⇒ b2 = 7

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