Question

If PQ is a double ordinate of the hyperbola ${\displaystyle \frac{x^2}{a^2}}-{\displaystyle \frac{y^2}{b^2}}=1$ such that OPQ is an equilateral triangle, O being the centre of the hyperbola, then the eccentricity e of the hyperbola satisfies,
A. $1<e<{\displaystyle \frac{2}{\sqrt{3}}}$
B. $e={\displaystyle \frac{2}{\sqrt{3}}}$
C. $e={\displaystyle \frac{\sqrt{3}}{2}}$
D. $e>{\displaystyle \frac{2}{\sqrt{3}}}$

Answer 1

- Hint: We will be using the concept of hyperbola to solve the problem. We will first use the parametric form of a point on hyperbola to find a general point on hyperbola then we will use the given condition that the triangle is equilateral to find a relation between a and b and then we will be using the concept of eccentricity double ordinate to further simplify the problem.

Complete step-by-step solution -

We have been given a hyperbola ${\displaystyle \frac{x^2}{a^2}}-{\displaystyle \frac{y^2}{b^2}}=1$ and that PQ is a double ordinate such that OPQ is an equilateral triangle.

Now, we have been given that $\mathrm{\Delta }OPQ$ is an equilateral triangle therefore, the $\mathrm{\angle }POQ=60{}^{\circ }$.
Now, since PQ is a double ordinate therefore the $\mathrm{\angle }POMand\mathrm{\angle }MOQ$ are equal due to symmetry. So, we have,
$\mathrm{\angle }POM=\mathrm{\angle }MOQ={\displaystyle \frac{60}{2}}=30{}^{\circ }$
Now, we take the coordinate of P in parametric form as $(a\sec \theta ,b\tan \theta )$. Also, the coordinate of O origin is (0, 0).
Therefore, the slope OP is,
${\displaystyle \frac{b\tan \theta -0}{a\sec \theta -0}}$
From the formula ${\displaystyle \frac{y_2-y_1}{x_2-x_1}}=m$.
Also, the slope is equal to tan 30 as we have been shown above. Therefore,
$\begin{array}{rl}& {\displaystyle \frac{b\tan \theta }{a\sec \theta }}=\tan 30{}^{\circ }\\ & {\displaystyle \frac{b\tan \theta }{a\sec \theta }}={\displaystyle \frac{1}{\sqrt{3}}}\end{array}$
Now, we will use the identities,
$\begin{array}{rl}& \tan \theta ={\displaystyle \frac{\sin \theta }{\cos \theta }}\\ & \sec \theta ={\displaystyle \frac{1}{\cos \theta }}\\ & {\displaystyle \frac{b}{a}}{\displaystyle \frac{\sin \theta }{\cos \theta }}\times \cos \theta ={\displaystyle \frac{1}{\sqrt{3}}}\\ & \sin \theta ={\displaystyle \frac{a}{b}}\times {\displaystyle \frac{1}{\sqrt{3}}}\\ & \Rightarrow \text{cosec}\theta ={\displaystyle \frac{b\sqrt{3}}{a}}\end{array}$
Now, we will square both sides to simplify it,
$\text{cose}{\text{c}}^2\theta ={\displaystyle \frac{3b^2}{a^2}}...........\left(1\right)$
Now, we know that the eccentricity of hyperbola is given by,
$\begin{array}{rl}& b^2=a^2(e^2-1)\\ & {\displaystyle \frac{b^2}{a^2}}=e^2-1\\ & {\displaystyle \frac{b^2}{a^2}}+1=e^2\end{array}$
We will substitute the value of ${\displaystyle \frac{b^2}{a^2}}$ from (1),
$\begin{array}{rl}& {\displaystyle \frac{\text{cose}{\text{c}}^2\theta }{3}}+1=e^2\\ & =\text{cose}{\text{c}}^2\theta =3(e^2-1)..........\left(2\right)\end{array}$
Now, we know that the range of $\text{cosec}\theta$ is $(-\mathrm{\infty },-1]\cup [1,\mathrm{\infty })$.
So, the range of $\text{cose}{\text{c}}^2\theta$ is $\text{cose}{\text{c}}^2\theta \ge 1$.
Now, we will use this in (2), where $\text{cose}{\text{c}}^2\theta =3(e^2-1)$.
$\begin{array}{rl}& \Rightarrow 3(e^2-1)\ge 1\\ & e^2-1\ge {\displaystyle \frac{1}{3}}\\ & e^2\ge {\displaystyle \frac{1}{3}}+1\\ & e^2\ge {\displaystyle \frac{4}{3}}\\ & e^2>\sqrt{{\displaystyle \frac{4}{3}}}\\ & e>\pm {\displaystyle \frac{2}{\sqrt{3}}}\end{array}$
We will ignore the negative inequality since e > 1 for hyperbola. Therefore,
$e>{\displaystyle \frac{2}{\sqrt{3}}}$ is the answer.
Hence, option (D) is correct.

Note: To solve these types of questions one must have a basic understanding of the concepts of hyperbola like double ordinate also it is important to note how we have used the condition of equilateral triangle to solve the problem.