Question

Let $H \dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ ,where $[a\text{ }>\text{ }b\text{ }>\text{ }0$, be a hyperbola in the $\text{xy} $ plane whose conjugate axis $\text{LM}$ subtends an angle of ${{60}^{\circ }}$at one of its vertices $\text{N}$. Let the area of the triangle $LMN$ be 4√3.
\[\begin{matrix}
   List-I & List-II \\
\text{P}.\text{ The length of the conjugate axis of}~\text{H} & \text{ 1}.~\text{8 } \\
\text{Q}.\text{ The eccentricity of}~\text{H}~\text{is } & 2.\dfrac{4}{\sqrt{3}} \\
\text{R}.\text{ The distance between the foci of}~\text{H}~\text{is} & 3.\dfrac{2}{\sqrt{3}} \\
\text{S}.\text{ The length of the latus rectum of}~\text{H}~\text{is} & \text{1}.~\text{4} \\
\end{matrix}\]
The correct option is
A.$P\to 4,Q\to 2,R\to 1,S\to 3$ \[\]
B.$P\to 4,Q\to 3,R\to 1,S\to 2$\[\]
C.$P\to 4,Q\to 1,R\to 3,S\to 2$\[\]
D.$P\to 3,Q\to 4,R\to 2,S\to 1$\[\]

Answer 1

Hint: Draw a figure to visualize the hyperbola, its conjugate axis which subtends the angle. Take the tangent of the angle $\angle LNO$ in the triangle LMN and find a relation $a=b\sqrt{3}$. Then use the area of a triangle $\dfrac{1}{2}\times $base$\times $perpendicular$=4\sqrt{3}$ to find $a$ and $b$. The formulas for length of conjugate axis, eccentricity, distance between the foci, length of latus rectum are $2b,\sqrt{1+\dfrac{{{b}^{2}}}{{{a}^{2}}}},2ae,\dfrac{2{{b}^{2}}}{a}$ respectively. \[\]

Complete step-by-step answer:
The given equation of parabola is \[H:\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1\]
We see from the figure that the hyperbola cuts $x$-axis at the points with coordinates $(a,0)$ and $(-a,0)$ symmetrically as the given hyperbola is centered at origin. So we denote one of the points of intersection say $(a,0)$ as$N$. We can denote to $(-a,0)$ too but the result will be the same.
Here the conjugate axis $LM$ lies on $x$-axis and subtends an angle of ${{60}^{\circ }}$ at the point $N$and the triangle $\Delta LMN$ is formed. As the hyperbola is also symmetric also along the transverse $y$–axis , $LM=2OL=2b$.As the hyperbola is symmetrical along $x$-axis$\angle LNO=\angle MNO=\dfrac{{{60}^{\circ }}}{2}={{30}^{\circ }}=\theta $ (say). Then
\[\begin{align}
  & \tan \theta =\dfrac{b}{a} \\
 & \Rightarrow \tan {{30}^{\circ }}=\dfrac{b}{a} \\
 & \Rightarrow a=b\sqrt{3} \\
\end{align}\]
  Using the formula of a area of a triangle with base and height of perpendicular
\[\begin{align}
  & \dfrac{1}{2}LM\cdot OL=4\sqrt{3} \\
 & \Rightarrow \dfrac{1}{2}2b\cdot b\sqrt{3}=4\sqrt{3} \\
 & \Rightarrow b=2 \\
\end{align}\]
So $a=2\sqrt{3}$
Now we calculate eccentricity using the formula,

\[e=\sqrt{1+\dfrac{{{b}^{2}}}{{{a}^{2}}}}=\sqrt{1+\dfrac{{{2}^{2}}}{{{\left( 2\sqrt{3} \right)}^{2}}}}=\dfrac{2}{\sqrt{3}}\]
Now we shall check each of the given options .
P. The length of the conjugate axis is the length of $LM=2b=4$. So $P\to 4$\[\]
Q. The eccentricity of H is $e=\dfrac{2}{\sqrt{3}}$. So $ \text{Q}\to \text{3}$\[\]
R. The distance between the foci is $2ae=2.2\sqrt{3}.\dfrac{2}{\sqrt{3}}$. So $ \text{R}\to \text{3}$ \[\]
S. The length of the latus rectum is $\dfrac{2{{b}^{2}}}{a}=\dfrac{2{{\left( 2 \right)}^{2}}}{2\sqrt{3}}=\dfrac{4}{\sqrt{3}}$. So $ \text{S}\to \text{3}$.\[\]

So, the correct answer is “Option B”.

Note: While solving problem of conic section like this type, we need to be careful of confusion between the formulas hyperbola from ellipse which is given by $e=\sqrt{1-\dfrac{{{b}^{2}}}{{{a}^{2}}}}$. We also note that thoughts have more than one eccentricity. The value of eccentricity for hyperbola is greater than 1 and for ellipse less than 1 .