Question

If the foci of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ coincide with the foci of the ellipse $\frac{x^2}{49} + \frac{y^2}{36} = 1$, then the value of $a^2 + b^2$ is equal to

• A. 25
• B. 36
• C. 85
• D. 169
• E. 13

Hint:

For confocal conics, the property \( a_h^2 e_h^2 = a_e^2 e_e^2 \) holds. For these standard horizontal forms, it simply means \( a^2 + b^2 \) for the hyperbola equals \( A^2 - B^2 \) for the ellipse.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
The foci of both an ellipse and a hyperbola centered at the origin are located at \((\pm c, 0)\).
Since their foci coincide, the value of \(c^2\) must be the same for both.
Step 2: Key Formula or Approach:
For an ellipse \(\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1\), the foci distance squared is \(c^2 = A^2 - B^2\).
For a hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the foci distance squared is \(c^2 = a^2 + b^2\).
Equate the two expressions for \(c^2\).
Step 3: Detailed Explanation:
Find \(c^2\) for the ellipse:
The equation is \(\frac{x^2}{49} + \frac{y^2}{36} = 1\), so \(A^2 = 49\) and \(B^2 = 36\).
\[ c^2 = A^2 - B^2 = 49 - 36 = 13 \] Find \(c^2\) for the hyperbola:
The equation is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\).
\[ c^2 = a^2 + b^2 \] Since the foci coincide, set them equal:
\[ a^2 + b^2 = 13 \] Step 4: Final Answer:
The value of \(a^2 + b^2\) is 13.