Question

The length of the transverse axis of a hyperbola is \( 2\cos \alpha \). The foci of the hyperbola are the same as that of the ellipse \( 9x^2 + 16y^2 = 144 \). The equation of the hyperbola is:

• A. \( \frac{x^2}{\cos^2 \alpha} - \frac{y^2}{7 - \cos^2 \alpha} = 1 \)
• B. \( \frac{x^2}{\cos^2 \alpha} - \frac{y^2}{7 + \cos^2 \alpha} = 1 \)
• C. \( \frac{x^2}{1 + \cos^2 \alpha} - \frac{y^2}{7 - \cos^2 \alpha} = 1 \)
• D. \( \frac{x^2}{1 + \cos^2 \alpha} - \frac{y^2}{7 + \cos^2 \alpha} = 1 \)
• E. \( \frac{x^2}{\cos^2 \alpha} - \frac{y^2}{5 - \cos^2 \alpha} = 1 \)

Hint:

In confocal problems, always calculate $c^2$ (the square of the distance from center to focus) first. It acts as the "bridge" between the two different conic equations.

Answer 1

The Correct Option is A

Concept: For two conics to be confocal (sharing the same foci), the distance \( c = ae \) must be identical for both. For an ellipse, \( c^2 = a^2 - b^2 \). For a hyperbola, \( c^2 = a^2 + b^2 \).

Step 1: Find the foci of the given ellipse.
Divide \( 9x^2 + 16y^2 = 144 \) by 144: \[ \frac{x^2}{16} + \frac{y^2}{9} = 1 \] Here, \( a^2 = 16 \) and \( b^2 = 9 \). The distance to the focus \( c \) is: \[ c^2 = a^2 - b^2 = 16 - 9 = 7 \]

Step 2: Find the parameters of the hyperbola.
Let the hyperbola equation be \( \frac{x^2}{A^2} - \frac{y^2}{B^2} = 1 \). Length of transverse axis \( 2A = 2\cos\alpha \), so \( A = \cos\alpha \). \[ A^2 = \cos^2\alpha \] Since the foci are the same, the \( c^2 \) for the hyperbola must also be 7. For a hyperbola, \( c^2 = A^2 + B^2 \). \[ 7 = \cos^2\alpha + B^2 \] \[ B^2 = 7 - \cos^2\alpha \]

Step 3: Form the equation.
\[ \frac{x^2}{\cos^2\alpha} - \frac{y^2}{7 - \cos^2\alpha} = 1 \]