Question:

If $PQ$ is a double ordinate of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ such that $\Delta OPQ$ is equilateral, $O$ being the centre. Then the eccentricity e satisfies

Updated On: Apr 27, 2024
  • $1 < e < \frac{2}{\sqrt{3}}$
  • $e=\frac{2}{\sqrt{2}}$
  • $e=\frac{\sqrt{3}}{2}$
  • $e>\frac{2}{\sqrt{3}}$
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is D

Solution and Explanation

$\because \Delta OPQ$ is equilateral, $OP = PQ$ $\Rightarrow a^{2}\,sec^{2}\,\theta+b^{2}\,tan^{2}\,\theta=\left(2b\,tan\,\theta\right)^{2}$ $\Rightarrow a^{2}\,sec^{2}\,\theta+3b^{2}\,tan^{2}\,\theta$ $\Rightarrow sin^{2}\,\theta= \frac{a^{2}}{3b^{2}}$ Now, $sin^{2}\,\theta <1$ $\Rightarrow \frac{a^{2}}{3b^{2}} >1$ $\Rightarrow \frac{b^{2}}{a^{2}}>\frac{1}{3} \Rightarrow 1+\frac{b^{2}}{a^{2}} > \frac{4}{3} \Rightarrow e^{2}>\frac{4}{3} \Rightarrow e> \frac{2}{\sqrt{3}}$
Was this answer helpful?
0
0

Concepts Used:

Hyperbola

Hyperbola is the locus of all the points in a plane such that the difference in their distances from two fixed points in the plane is constant.

Hyperbola is made up of two similar curves that resemble a parabola. Hyperbola has two fixed points which can be shown in the picture, are known as foci or focus. When we join the foci or focus using a line segment then its midpoint gives us centre. Hence, this line segment is known as the transverse axis.

Hyperbola