Question

If the latus rectum of a hyperbola is 8 and eccentricity is $\dfrac{3}{\sqrt{5}}$ then the equation of the hyperbola is?
(a) $4{{x}^{2}}-5{{y}^{2}}=100$
(b) $5{{x}^{2}}-4{{y}^{2}}=100$
(c) $4{{x}^{2}}+5{{y}^{2}}=100$
(d) $5{{x}^{2}}+4{{y}^{2}}=100$

Answer 1

Hint:Assume the general equation of hyperbola as $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$. Here, a is the axis of the hyperbola and b is the conjugate axis. Apply the formula: $e=\sqrt{1+\dfrac{{{b}^{2}}}{{{a}^{2}}}}$, where ‘e’ is the eccentricity, to form a equation in a and b. Use another relation given by: $l=\dfrac{2{{b}^{2}}}{a}$, where ‘l’ is the length of the latus rectum, to form another equation in a and b. Now, solve the two equations to get the value of a and b. Finally, substitute the value of a and b in the general equation of hyperbola.

Complete step-by-step answer:
Now, let us assume the equation of hyperbola as $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$, where a is the axis of the hyperbola and b is the conjugate axis.

It is given that the length of the latus rectum is 8. Therefore, applying the formula for length of latus rectum of hyperbola, we get,
$\begin{align}
  & l=\dfrac{2{{b}^{2}}}{a} \\
 & \Rightarrow 8=\dfrac{2{{b}^{2}}}{a} \\
 & \Rightarrow \dfrac{{{b}^{2}}}{a}=4..................(i) \\
\end{align}$
Also, it is given that eccentricity is $\dfrac{3}{\sqrt{5}}$. Therefore, applying the formula for eccentricity of hyperbola, we get,
$\begin{align}
  & e=\sqrt{1+\dfrac{{{b}^{2}}}{{{a}^{2}}}} \\
 & \Rightarrow \dfrac{3}{\sqrt{5}}=\sqrt{1+\dfrac{{{b}^{2}}}{{{a}^{2}}}} \\
\end{align}$
On squaring both sides, we get,
$\begin{align}
  & \dfrac{9}{5}=1+\dfrac{{{b}^{2}}}{{{a}^{2}}} \\
 & \Rightarrow \dfrac{{{b}^{2}}}{{{a}^{2}}}=\dfrac{9}{5}-1 \\
 & \Rightarrow \dfrac{{{b}^{2}}}{{{a}^{2}}}=\dfrac{4}{5}.............................(ii) \\
\end{align}$
Substituting the value of $\dfrac{{{b}^{2}}}{a}$ from equation (i) in equation (ii), we get,
$\begin{align}
  & \dfrac{{{b}^{2}}}{a}\times \dfrac{1}{a}=\dfrac{4}{5} \\
 & \Rightarrow \dfrac{4}{a}=\dfrac{4}{5} \\
 & \Rightarrow \dfrac{1}{a}=\dfrac{1}{5} \\
 & \Rightarrow a=5 \\
\end{align}$
Substituting (a = 5) in equation (i), we get,
$\begin{align}
  & {{b}^{2}}=5\times 4=20 \\
 & \Rightarrow b=\sqrt{20} \\
\end{align}$
Now, substituting the value of a and b in the general equation of hyperbola, we get,
$\begin{align}
  & \dfrac{{{x}^{2}}}{{{5}^{2}}}-\dfrac{{{y}^{2}}}{{{\left( \sqrt{20} \right)}^{2}}}=1 \\
 & \Rightarrow \dfrac{{{x}^{2}}}{25}-\dfrac{{{y}^{2}}}{20}=1 \\
\end{align}$
Multiplying both sides by 100, we get,
$\begin{align}
  & 100\times \left( \dfrac{{{x}^{2}}}{25}-\dfrac{{{y}^{2}}}{20} \right)=100 \\
 & \Rightarrow 4{{x}^{2}}-5{{y}^{2}}=100 \\
\end{align}$
Hence, option (a) is the correct answer.

Note: One should not get confused in the general equation of hyperbola and ellipse that has a difference of only minus sign in the middle. Also note that the eccentricity of a hyperbola is always greater than 1 while that of ellipse is less than 1. As you can see that we have multiplied both sides of the equation with 100. This is because of the options given. You may note that 100 is the L.C.M of 25 and 20.