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
- A \(4\)
- B \(5\)
- C \(8\)
- D \(9\)
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…
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