Question

PQ is double ordinate of hyperbola $${\displaystyle \frac{x^2}{a^2}}-{\displaystyle \frac{y^2}{b^2}}=1$$ such that OPQ is an equilateral triangle, O being the center of hyperbola, where the eccentricity of the hyperbola $$e$$ satisfy,$$\sqrt{3}e>k$$, then the value of k is

Answer 1

Hint: Here we have to find the value of eccentricity of the hyperbola. We will first write the polar coordinates of the point P and point Q. Here it is given that the OPQ is an equilateral triangle. O is the center of the hyperbola; all the angles of the triangle will be $${60}^{\circ }$$. Then we will apply the distance formula and eccentricity formula for hyperbola to get the value of k.

Complete step-by-step answer:
We will draw the figure first with appropriate data given.

It is given that O is the center of the hyperbola and PQ is the double ordinate of the hyperbola. So the coordinate of point P $$=\left(a\sec \theta ,b\tan \theta \right)$$ and coordinate of Q $$=\left(a\sec \theta ,-b\tan \theta \right)$$
Since $$\mathrm{\Delta }POQ$$ is an equilateral triangle. So the length of all sides is equal. Thus, $$PO=QO=PQ$$.
We will find the distance between these two points, $$\left(a\sec \theta ,-b\tan \theta \right)$$ and $$\left(a\sec \theta ,b\tan \theta \right)$$.
Now we will replace $$x_1$$ with $$a\sec \theta$$, $$y_1$$ with $$-b\tan \theta$$, $$x_2$$ with $$a\sec \theta$$ and $$y_2$$ with $$b\tan \theta$$ in the distance formula $$\sqrt{\left[{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2\right]}$$.
Distance of the point $$\left(a\sec \theta ,-b\tan \theta \right)$$ from $$\left(a\sec \theta ,b\tan \theta \right)$$$$=\sqrt{\left[{\left(a\sec \theta -a\sec \theta \right)}^2+{\left(b\tan \theta -(-b\tan \theta )\right)}^2\right]}$$
Subtracting the terms and applying exponents on the bases, we get
Distance of the point $$\left(a\sec \theta ,-b\tan \theta \right)$$from $$\left(a\sec \theta ,b\tan \theta \right)$$ $$=\sqrt{0+4b^2{\tan }^2\theta }$$ …………..$$\left(1\right)$$
Similarly, we will find the distance between the points $$\left(0,0\right)$$ and $$\left(a\sec \theta ,b\tan \theta \right)$$.
Now we will replace $$x_1$$ with 0, $$y_1$$ with 0, $$x_2$$ with $$a\sec \theta$$ and $$y_2$$ with $$b\tan \theta$$ in the distance formula $$\sqrt{\left[{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2\right]}$$.
Distance of the point $$\left(0,0\right)$$ from $$\left(a\sec \theta ,b\tan \theta \right)$$ $$=\sqrt{\left[{\left(a\sec \theta -0\right)}^2+{\left(b\tan \theta -0\right)}^2\right]}$$
Applying exponents on the bases, we get
Distance of point $$\left(0,0\right)$$ from $$\left(a\sec \theta ,b\tan \theta \right)$$ $$=\sqrt{a^2{\sec }^2\theta +b^2{\tan }^2\theta }$$………….$$\left(2\right)$$
Squaring and equating equation $$\left(1\right)$$ and equation $$\left(2\right)$$, we get
$${\left(\sqrt{0+4b^2{\tan }^2\theta }\right)}^2={\left(\sqrt{a^2{\sec }^2\theta +b^2{\tan }^2\theta }\right)}^2$$
Simplifying the equation, we get
$\Rightarrow$ $$4a^2{\tan }^2\theta =a^2{\sec }^2\theta +a^2{\tan }^2\theta$$
Adding and subtracting like terms, we get
$\Rightarrow$ $$3b^2{\tan }^2\theta =a^2{\sec }^2\theta$$
On further simplification, we get
$\Rightarrow$ $${\displaystyle \frac{b^2}{a^2}}={\displaystyle \frac{{\sec }^2\theta }{3{\tan }^2\theta }}$$
Breaking the terms, we get
$\Rightarrow$ $${\displaystyle \frac{b^2}{a^2}}={\displaystyle \frac{{\displaystyle \frac{1}{{\cos }^2\theta }}}{{\displaystyle \frac{3{\sin }^2\theta }{{\cos }^2\theta }}}}={\displaystyle \frac{1}{3{\sin }^2\theta }}$$
We know the square of eccentricity of hyperbola is equal to $$1+{\displaystyle \frac{b^2}{a^2}}$$ i.e. $$e^2=1+{\displaystyle \frac{b^2}{a^2}}$$
We will put the value of $${\displaystyle \frac{b^2}{a^2}}$$ that we have calculated, in the above formula.
$\Rightarrow$ $$e^2=1+{\displaystyle \frac{1}{3{\sin }^2\theta }}$$
Simplifying the terms further, we get
$\Rightarrow$ $$e^2={\displaystyle \frac{1+4{\tan }^2\theta }{3{\tan }^2\theta }}$$
Rewriting the equation, we get
$\Rightarrow$ $${\displaystyle \frac{1}{3\left(e^2-1\right)}}={\sin }^2\theta$$
We know that $${\sin }^2\theta <1$$.
Therefore,
$\Rightarrow$ $${\displaystyle \frac{1}{3\left(e^2-1\right)}}<1$$
Using inequality property, we get
$\Rightarrow$ $$3\left(e^2-1\right)>1$$
Simplifying the terms, we get
$\Rightarrow$ $$e^2-1>{\displaystyle \frac{1}{3}}$$
Adding 1 on both sides, we get
$$\begin{gathered} \Rightarrow e^2-1+1>{\displaystyle \frac{1}{3}}+1 \\ \Rightarrow e^2>{\displaystyle \frac{4}{3}} \end{gathered}$$
Taking square root on both sides, we get
$$\begin{gathered} \Rightarrow \sqrt{e^2}>\sqrt{{\displaystyle \frac{4}{3}}} \\ \Rightarrow e>{\displaystyle \frac{2}{\sqrt{3}}} \\ \Rightarrow \sqrt{3}e>2 \end{gathered}$$
Thus, the value of $$k$$ is 2.

Note: We have found out the required answer using the eccentricity of hyperbola. Eccentricity is a measure that tells how much a conic section differs from being circular. Conic section includes the circle, hyperbola, parabola etc. Different conic sections have different eccentricity. For example, a circle has an eccentricity 0 whereas hyperbola has an eccentricity usually greater than 1. It therefore becomes very important to keep in mind the formula of eccentricity for different conic sections.