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Conic Sections — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsConic Sections2 Marks
MCQ
A hyperbola, centred at the origin, has transverse axis 2 a . If it passes through a given point $\left(x_1, y_1\right)$, then its eccentricity is
  • A
    $\sqrt{\frac{x_1^2-y_1^2- a ^2}{x_1^2-y_1^2}}$
  • ✓
    $\sqrt{\frac{ a ^2-x_1^2-y_1^2}{ a ^2-x_1^2}}$
  • C
    $\sqrt{\frac{ a ^2+x_1^2+y_1^2}{ a ^2-x_1^2}}$
  • D
    None of these
✓

Answer

Correct option: B.
$\sqrt{\frac{ a ^2-x_1^2-y_1^2}{ a ^2-x_1^2}}$
(B)
Equation of hyperbola passes through
$\begin{array}{l}\left(x_1, y_1\right) \\\Rightarrow \frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}=1 \Rightarrow \frac{x_1^2}{a^2}-1=\frac{y_1^2}{b^2} \\
\Rightarrow \frac{x_1^2-a^2}{a^2}=\frac{y_1^2}{b^2} \Rightarrow \frac{b^2}{a^2}=\frac{y_1^2}{x_1^2-a^2}\end{array}$
Now, $\frac{b^2}{a^2}=e^2-1$
$\begin{array}{l}\Rightarrow \frac{y_1^2}{x_1^2-a^2}=e^2-1 \\
\Rightarrow e^2=\frac{y_1^2+\left(x_1^2-a^2\right)}{x_1^2-a^2} \\
\Rightarrow e=\sqrt{\frac{x_1^2-a^2+y_1^2}{x_1^2-a^2}}=\sqrt{\frac{a^2-x_1^2-y_1^2}{a^2-x_1^2}}\end{array}$

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