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TS EAMCET · Maths · Hyperbola

If the focii of the ellipse \(\frac{x^2}{25}+\frac{y^2}{16}=1\) and the hyperbola \(\frac{x^2}{4}-\frac{y^2}{b^2}=1\) coincide, then \(b^2\) is equal to

  1. A \(4\)
  2. B \(5\)
  3. C \(8\)
  4. D \(9\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(5\)

Step-by-step Solution

Detailed explanation

Given, equation of ellipse is \[ \frac{x^2}{25}+\frac{y^2}{16}=1 \] and equation of hyperbola is \[ \frac{x^2}{4}-\frac{y^2}{b^2}=1 \] eccentricity of ellipse…