Question

Let \(0 < {\rm{\theta }} < \frac{{\rm{\pi }}}{2}\). If the eccentricity of the hyperbola \(\frac{{{{\rm{x}}^2}}}{{{\rm{co}}{{\rm{s}}^2}{\rm{\theta }}}} - \frac{{{{\rm{y}}^2}}}{{{\rm{si}}{{\rm{n}}^2}{\rm{\theta }}}} = 1\) is greater than 2, then the length of its latus rectum lies in the interval:
1. (3, ∞)
2. \(\left( {\frac{3}{2},{\rm{\;}}2} \right]\)
3. (2, 3]
4. \(\left( {1,{\rm{\;}}\frac{3}{2}} \right]\)

Answer 1

Correct Answer - Option 1 : (3, ∞)

From question, the hyperbola is given by the equation:

\(\frac{{{{\rm{x}}^2}}}{{{\rm{co}}{{\rm{s}}^2}{\rm{\theta }}}} - \frac{{{{\rm{y}}^2}}}{{{\rm{si}}{{\rm{n}}^2}{\rm{\theta }}}} = 1\) 

Now, the general equation of hyperbola is:

\(\frac{{{{\rm{x}}^2}}}{{{{\rm{a}}^2}}} - \frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}} = 1\) 

On comparing, general equation and given equation,

⇒ a² = cos² θ

⇒ b² = sin² θ

Given, the eccentricity of the hyperbola is greater than 2.

The eccentricity of the hyperbola is given by the formula:

\({\rm{e}} = \sqrt {1 + \frac{{{{\sin }^2}{\rm{\theta }}}}{{{{\cos }^2}{\rm{\theta }}}}}\) 

Now,

⇒ e > 2

\(\Rightarrow \sqrt {1 + \frac{{{{\sin }^2}{\rm{\theta }}}}{{{{\cos }^2}{\rm{\theta }}}}} > 2\) 

\(\because \left[ \tan \text{ }\!\!\theta\!\!\text{ }=\frac{\sin \text{ }\!\!\theta\!\!\text{ }}{\cos \text{ }\!\!\theta\!\!\text{ }} \right]\) 

\(\Rightarrow \sqrt{1+{{\tan }^{2}}\text{ }\!\!\theta\!\!\text{ }}>2\) 

On squaring both sides,

⇒ 1 + tan² θ > 4

⇒ tan² θ > 3

Taking square root on both sides,

\(\Rightarrow \tan \text{ }\!\!\theta\!\!\text{ }>\sqrt{3}\) 

Let’s take tan θ as x for easy calculation.

\(\Rightarrow \text{x}>\sqrt{3}\Rightarrow \left| \text{x} \right|>\sqrt{3}\Rightarrow \text{x}\in \left( -\infty ,\text{ }\!\!~\!\!\text{ }-\sqrt{3} \right)\cup \left( \sqrt{3},\text{ }\!\!~\!\!\text{ }\infty \right)\)

From question, \(\text{ }\!\!\theta\!\!\text{ }\in \left[ 0,\text{ }\!\!~\!\!\text{ }\frac{\text{ }\!\!\pi\!\!\text{ }}{2} \right]\)

\(\Rightarrow \tan \text{ }\!\!\theta\!\!\text{ }\in \left( \sqrt{3},\text{ }\!\!~\!\!\text{ }\infty \right)\) 

\(\because \tan \frac{\text{ }\!\!\pi\!\!\text{ }}{3}=\sqrt{3}~\text{and}~\tan \frac{\text{ }\!\!\pi\!\!\text{ }}{2}=\infty\) 

\(\therefore \text{ }\!\!\theta\!\!\text{ }\in \left( \frac{\text{ }\!\!\pi\!\!\text{ }}{3},\text{ }\!\!~\!\!\text{ }\frac{\text{ }\!\!\pi\!\!\text{ }}{2} \right)\) 

From question, we need to find the interval of the length of its latus rectum. So, we need to find the maximum value and minimum value.

The length of the latus rectum of hyperbola is given by the formula:

\(\text{LR}=\frac{2{{\text{b}}^{2}}}{\text{a}}\) 

On substituting the values,

\(\Rightarrow \text{LR}=\frac{2\left( {{\sin }^{2}}\text{ }\!\!\theta\!\!\text{ } \right)}{\cos \text{ }\!\!\theta\!\!\text{ }}\) 

⇒ LR = 2 sin θ.tan θ

The minimum length of the latus rectum \(\left[ \text{ }\!\!\theta\!\!\text{ }\in \frac{\text{ }\!\!\pi\!\!\text{ }}{3} \right]\) is:

\(\Rightarrow \text{L}{{\text{R}}_{\text{min}}}=2\sin \left( \frac{\text{ }\!\!\pi\!\!\text{ }}{3} \right).\tan \left( \frac{\text{ }\!\!\pi\!\!\text{ }}{3} \right)\) 

∵ \(\left[ \begin{matrix} \sin \left( \frac{\text{ }\!\!\pi\!\!\text{ }}{3} \right)=\frac{\sqrt{3}}{2} \\ \tan \left( \frac{\text{ }\!\!\pi\!\!\text{ }}{3} \right)=\sqrt{3} \\ \end{matrix} \right]\)

\(\Rightarrow \text{L}{{\text{R}}_{\text{min}}}=2\left( \frac{\sqrt{3}}{2} \right)\sqrt{3}\) 

∴ LR_min = 3

The maximum length of the latus rectum \(\left[ \text{ }\!\!\theta\!\!\text{ }\in \frac{\text{ }\!\!\pi\!\!\text{ }}{2} \right]\) is:

\(\Rightarrow \text{L}{{\text{R}}_{\text{max}}}=2\sin \left( \frac{\text{ }\!\!\pi\!\!\text{ }}{2} \right).\tan \left( \frac{\text{ }\!\!\pi\!\!\text{ }}{2} \right)\) 

\(\because \left[ \begin{matrix} \sin \left( \frac{\text{ }\!\!\pi\!\!\text{ }}{2} \right)=1 \\ \tan \left( \frac{\text{ }\!\!\pi\!\!\text{ }}{2} \right)=\infty \\ \end{matrix} \right]\) 

⇒ LR_max = 2(1)(∞)

∴ LR_max = ∞

Now, the interval of the latus rectum is (3, ∞).