AP EAMCET · Maths · Ellipse
Let \(\mathrm{X}\) - axis be the transverse axis and \(\mathrm{Y}\) - axis be the conjugate axis of a hyperbola \(\mathrm{H}\). Let the eccentricity of \(\mathrm{H}\) be the reciprocal of the eccentricity of the ellipse \(\frac{x^2}{4}+\frac{y^2}{2}=1\). If \((5,4)\) is a point on \(H\), then the length of the transverse axis of \(\mathrm{H}\) is
- A \(2 \sqrt{2}\)
- B 4
- C 6
- D 10
Answer & Solution
Correct Answer
(C) 6
Step-by-step Solution
Detailed explanation
Eccentricity of ellipse \(\frac{x^2}{4}+\frac{y^2}{2}\) is: \(e^{\prime}=\sqrt{1-\frac{2}{4}}=\sqrt{1-\frac{1}{2}}=\frac{1}{\sqrt{2}}\) Now eccentricity of hyperbola:…
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