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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Hyperbola

JEE Main · Mathematics · Hyperbola

If e_1 is the eccentricity of the hyperbola x= θ , y= 2 θ and e_2...

If e1e_1 is the eccentricity of the hyperbola x=sec⁡θx=\sec \theta, y=2tan⁡θy=\sqrt{2} \tan \theta and e2e_2 is the eccentricity of the hyperbola x=2sec⁡θ,y=tan⁡θx=\sqrt{2} \sec \theta, y=\tan \theta, then e22e12=\frac{e_2^2}{e_1^2}=

Options

  1. A1
  2. B2
  3. C12\frac{1}{2}
  4. D14\frac{1}{4}

Correct answer

C. 12\frac{1}{2}

Step-by-step solution

Acc. to given parametric coordinates e12=(12+(2)212)2\mathrm{e}_1^2=\left(\sqrt{\frac{1^2+(\sqrt{2})^2}{1^2}}\right)^2 and e22((2)2+(1)2(2)2)2\mathrm{e}_2^2\left(\sqrt{\frac{(\sqrt{2})^2+(1)^2}{(\sqrt{2})^2}}\right)^2 Now e22e12=12\frac{\mathrm{e}_2^2}{\mathrm{e}_1^2}=\frac{1}{2}
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