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Conic Sections — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsConic Sections2 Marks
MCQ
If the eccentricities of the hyperbolas $\frac{x^2}{ a ^2}-\frac{y^2}{b^2}=1$ and $\frac{y^2}{b^2}-\frac{x^2}{ a ^2}=1$ be e and $e_1$, then $\frac{1}{e^2}+\frac{1}{e^2}=$
  • ✓
    1
  • B
    2
  • C
    3
  • D
    4
✓

Answer

Correct option: A.
1
(A)
Eccentricity of hyperbola $\frac{x^2}{ a ^2}-\frac{y^2}{b^2}=1$ is
$\begin{array}{l}e=\sqrt{1+\frac{b^2}{a^2}} \\\Rightarrow e^2=\frac{a^2+b^2}{a^2} \quad\ldots(i)\end{array}$
Eccentricity of hyperbola $\frac{y^2}{b^2}-\frac{x^2}{ a ^2}=1$ is
$\begin{array}{l} e_1=\sqrt{1+\frac{a^2}{b^2}} \\\Rightarrow e_1^2=\frac{b^2+a^2}{b^2}\quad\ldots(ii)\end{array}$
From (i) and (ii), we get
$\frac{1}{e_1^2}+\frac{1}{e^2}=1$

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