Question

The equation of the hyperbola in the standard form ( with transverse axis along the x-axis) having the length of the latus rectum$=9$ units and eccentricity $=\dfrac{5}{4}$ is
A. $\dfrac{{{x}^{2}}}{16}-\dfrac{{{y}^{2}}}{18}=1$
B. $\dfrac{{{x}^{2}}}{36}-\dfrac{{{y}^{2}}}{27}=1$
C. $\dfrac{{{x}^{2}}}{64}-\dfrac{{{y}^{2}}}{36}=1$
D. $\dfrac{{{x}^{2}}}{16}-\dfrac{{{y}^{2}}}{9}=1$

Answer 1

Hint: To solve this question we will use the standard equation of hyperbola $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$. We will first formulate two equations with the help of the formula of the length of the latus rectum and eccentricity. Then we will simplify both equations with the substitution method and determine the value of $a$ and $b$. We will then substitute the values of $a$ and $b$in the standard form of equation and get the required equation of hyperbola.

Formula Used: Eccentricity of hyperbola :$e=\sqrt{1+\dfrac{{{b}^{2}}}{{{a}^{2}}}}$.
Length of the latus rectum $=\dfrac{2{{b}^{2}}}{a}$.

Complete step by step solution: We have given length of the latus rectum as $9$ units and eccentricity $\dfrac{5}{4}$ units and we have to find the equation of the hyperbola in the standard form.
Using formula of the length of the latus rectum and eccentricity of hyperbola we will derive two equations,
Length of the latus rectum $=\dfrac{2{{b}^{2}}}{a}$
$\begin{align}
  & 9=\dfrac{2{{b}^{2}}}{a} \\
 & {{b}^{2}}=\dfrac{9a}{2}.....(i) \\
\end{align}$
Now eccentricity of hyperbola,
 $\begin{align}
  & e=\sqrt{1+\dfrac{{{b}^{2}}}{{{a}^{2}}}} \\
 & \dfrac{5}{4}=\sqrt{1+\dfrac{{{b}^{2}}}{{{a}^{2}}}} \\
\end{align}$
Squaring on both sides of the equation,
$\begin{align}
  & \dfrac{25}{16}=1+\dfrac{{{b}^{2}}}{{{a}^{2}}} \\
 & \dfrac{{{b}^{2}}}{{{a}^{2}}}=\dfrac{25}{16}-1 \\
 & \dfrac{{{b}^{2}}}{{{a}^{2}}}=\dfrac{9}{16} \\
 & {{b}^{2}}=\dfrac{9{{a}^{2}}}{16}....(ii)
\end{align}$
We will now substitute the equation (i) in (ii).
$\begin{align}
  & \dfrac{9a}{2}=\dfrac{9{{a}^{2}}}{16} \\
 & a=8
\end{align}$
Now value of $b$.
$\begin{align}
  & {{b}^{2}}=\dfrac{9a}{2} \\
 & {{b}^{2}}=\dfrac{9\times 8}{2} \\
 & {{b}^{2}}=36 \\
 & b=6
\end{align}$
Now we know that the standard equation of the hyperbola is $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$, so to find the standard equation of hyperbola we will substitute the value of $a$ and $b$ in it.
$\begin{align}
  & \dfrac{{{x}^{2}}}{{{8}^{2}}}-\dfrac{{{y}^{2}}}{{{6}^{2}}}=1 \\
 & \dfrac{{{x}^{2}}}{64}-\dfrac{{{y}^{2}}}{36}=1 \\
\end{align}$
> The standard equation of the hyperbola having the length of the latus rectum as $9$ units and eccentricity $\dfrac{5}{4}$ units is $\dfrac{{{x}^{2}}}{64}-\dfrac{{{y}^{2}}}{36}=1$

Option ‘C’ is correct

Note: We will consider the standard equation of hyperbola as $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ and its formula of eccentricity instead of the other standard equation of conjugate hyperbola $\dfrac{{{y}^{2}}}{{{b}^{2}}}-\dfrac{{{x}^{2}}}{{{a}^{2}}}=1$and its formula of eccentricity because it is mentioned in the question that transverse axis is along the x-axis and in conjugate hyperbola the transverse axis is along the y-axis.