Question

If the foci of the ellipse \( \frac{x^2}{25} + \frac{y^2}{b^2} = 1 \) and the hyperbola \( \frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25} \) are coincide, then the value of \( b^2 \) is

• A. 25
• B. 16
• C. 64
• D. 49

Hint:

For any conic section problem involving foci, the first step is to convert the equation to its standard form (i.e., = 1). Then, find the foci using the standard formulas: \( (\pm ae, 0) \) for an ellipse and \( (\pm AE, 0) \) for a hyperbola. Equate the foci coordinates and solve for the unknown.

Answer 1

The Correct Option is B

First, let's find the foci of the hyperbola. The equation is: \[ \frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25} \]
To get it into the standard form \( \frac{x^2}{A^2} - \frac{y^2}{B^2} = 1 \), multiply by 25:
\[ \frac{25x^2}{144} - \frac{25y^2}{81} = 1 \Rightarrow \frac{x^2}{144/25} - \frac{y^2}{81/25} = 1 \] For the hyperbola, \( A^2 = \frac{144}{25} \) and \( B^2 = \frac{81}{25} \). The eccentricity \( E \) of the hyperbola is given by \( E^2 = 1 + \frac{B^2}{A^2} \). \[ E^2 = 1 + \frac{81/25}{144/25} = 1 + \frac{81}{144} = 1 + \frac{9}{16} = \frac{25}{16} \Rightarrow E = \frac{5}{4} \] The foci of the hyperbola are at \( (\pm AE, 0) \). \[ A = \sqrt{\frac{144}{25}} = \frac{12}{5} \] \[ AE = \frac{12}{5} \times \frac{5}{4} = 3 \] So, the foci of the hyperbola are at \( (\pm 3, 0) \). Now, for the ellipse \( \frac{x^2}{25} + \frac{y^2}{b^2} = 1 \). Here, \( a^2 = 25 \), so \( a = 5 \). The foci of the ellipse are at \( (\pm ae, 0) \). Since the foci coincide, we have \( ae = 3 \). \[ 5e = 3 \Rightarrow e = \frac{3}{5} \] The relationship between \( a, b, \) and eccentricity \( e \) for an ellipse is \( b^2 = a^2(1 - e^2) \). \[ b^2 = 25 \left(1 - \left(\frac{3}{5}\right)^2\right) \] \[ b^2 = 25 \left(1 - \frac{9}{25}\right) = 25 \left(\frac{16}{25}\right) = 16 \] The value of \( b^2 \) is 16.