TS EAMCET · Maths · Hyperbola
If \(e_1\) is the eccentricity of the hyperbola \(x=\sec \theta\), \(y=\sqrt{2} \tan \theta\) and \(e_2\) is the eccentricity of the hyperbola \(x=\sqrt{2} \sec \theta, y=\tan \theta\), then \(\frac{e_2^2}{e_1^2}=\)
- A 1
- B 2
- C \(\frac{1}{2}\)
- D \(\frac{1}{4}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
Acc. to given parametric coordinates \(\mathrm{e}_1^2=\left(\sqrt{\frac{1^2+(\sqrt{2})^2}{1^2}}\right)^2\) and \(\mathrm{e}_2^2\left(\sqrt{\frac{(\sqrt{2})^2+(1)^2}{(\sqrt{2})^2}}\right)^2\) Now \(\frac{\mathrm{e}_2^2}{\mathrm{e}_1^2}=\frac{1}{2}\)
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