ExamBro
ExamBro
AP EAMCET · Maths · Hyperbola

If the product of eccentricities of the ellipse \(\frac{x^2}{16}+\frac{y^2}{b^2}=1\) and the hyperbola \(\frac{x^2}{9}-\frac{y^2}{16}=-1\) is 1 , then \(b^2=\)

  1. A \(\frac{12}{25}\)
  2. B \(144\)
  3. C \(25\)
  4. D \(\frac{144}{25}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{144}{25}\)

Step-by-step Solution

Detailed explanation

Given the ellipse \(\frac{x^2}{16}+\frac{y^2}{b^2}=1\) and hyperbola…