Question

If $e\,$and $e'$ are the eccentricities of the ellipse $5{{x}^{2}}+9{{y}^{2}}=45$ and the hyperbola $5{{x}^{2}}-4{{y}^{2}}=45$ respectively then $ee'=$.
A. $9$
B. $4$
C. $5$
D. $1$

Answer 1

Hint: To solve this question we will first write the given equation of ellipse as the general equation of ellipse $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ and then compare each other and find the value of ${{a}^{2}}$ and ${{b}^{2}}$. Then substituting values of ${{a}^{2}}$ and ${{b}^{2}}$ in the formula of eccentricity of ellipse we will find the value of $e\,$.
We will then take the equation of hyperbola and write it in the general equation of hyperbola $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ and then compare each other and derive the value of ${{a}^{2}}$ and ${{b}^{2}}$. Substituting values of ${{a}^{2}}$ and ${{b}^{2}}$ in the formula of eccentricity of hyperbola we will find the value of $e'$. We will then multiply the value of $e\,$and $e'$ and find the value of $ee'$.

Formula used: Eccentricity of hyperbola: $e=\sqrt{1+\dfrac{{{b}^{2}}}{{{a}^{2}}}}$.
Eccentricity of ellipse: $e=\sqrt{1-\dfrac{{{b}^{2}}}{{{a}^{2}}}}$.

Complete step by step solution: We are given that $e\,$and $e'$ are the eccentricities of the ellipse $5{{x}^{2}}+9{{y}^{2}}=45$ and hyperbola $5{{x}^{2}}-4{{y}^{2}}=45$ and we have to determine the value of $ee'$.
We know that the general equation of ellipse is $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$, so we will first write the given equation of ellipse $5{{x}^{2}}+9{{y}^{2}}=45$ in this form.
$5{{x}^{2}}+9{{y}^{2}}=45$
Dividing the equation by $45$ on both sides of the equation.
$\begin{align}
  & \dfrac{5{{x}^{2}}}{45}+\dfrac{9{{y}^{2}}}{45}=\dfrac{45}{45} \\
 & \dfrac{{{x}^{2}}}{9}+\dfrac{{{y}^{2}}}{5}=1......(i)
\end{align}$
We will now compare the equation (i) with the general equation of the ellipse and find the value of ${{a}^{2}}$ and ${{b}^{2}}$.
$\begin{align}
  & \dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=\dfrac{{{x}^{2}}}{9}+\dfrac{{{y}^{2}}}{5} \\
 & {{a}^{2}}=9 \\
 & {{b}^{2}}=5
\end{align}$
We will now calculate the eccentricity of ellipse $e\,$ with the help of the formula $e=\sqrt{1-\dfrac{{{b}^{2}}}{{{a}^{2}}}}$by substituting the value of ${{a}^{2}}$ and ${{b}^{2}}$.
\[\begin{align}
  & e=\sqrt{1-\dfrac{5}{9}} \\
 & e=\sqrt{\dfrac{4}{9}} \\
 & e=\dfrac{2}{3}
\end{align}\]
Now,
We know that the general equation of hyperbola is $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ so we will first write the given equation of hyperbola $5{{x}^{2}}-4{{y}^{2}}=45$ in this form.
 $5{{x}^{2}}-4{{y}^{2}}=45$
Dividing the equation by $45$ on both sides of the equation.
$\begin{align}
  & \dfrac{5{{x}^{2}}}{45}-\dfrac{4{{y}^{2}}}{45}=\dfrac{45}{45} \\
 & \dfrac{{{x}^{2}}}{9}-\dfrac{{{y}^{2}}}{\dfrac{45}{4}}=1......(ii)
\end{align}$
We will now compare the equation (ii) with the general equation of hyperbola and find the value of ${{a}^{2}}$ and ${{b}^{2}}$.
$\begin{align}
  & \dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=\dfrac{{{x}^{2}}}{9}-\dfrac{{{y}^{2}}}{\dfrac{45}{4}} \\
 & {{a}^{2}}=9 \\
 & {{b}^{2}}=\dfrac{45}{4}
\end{align}$
We will now find the eccentricity of hyperbola $e'$ using formula $e=\sqrt{1+\dfrac{{{b}^{2}}}{{{a}^{2}}}}$ by substituting the value of ${{a}^{2}}$ and ${{b}^{2}}$.
$\begin{align}
  & e'=\sqrt{1+\dfrac{\dfrac{45}{4}}{9}} \\
 & e'=\sqrt{1+\dfrac{45}{36}} \\
 & e'=\sqrt{\dfrac{81}{36}} \\
 & e'=\dfrac{9}{6} \\
 & e'=\dfrac{3}{2} \\
\end{align}$
We will now calculate the value of $ee'$.
$\begin{align}
  & ee'=\dfrac{2}{3}\times \dfrac{3}{2} \\
 & ee'=1 \\
\end{align}$

The value of $ee'$ is $ee'=1$ where $e\,$and $e'$ are the eccentricities of the ellipse $5{{x}^{2}}+9{{y}^{2}}=45$ and the hyperbola $5{{x}^{2}}-4{{y}^{2}}=45$. Hence the correct option is (D).

Note: The eccentricity of an ellipse should be always less than one while eccentricity of hyperbola is always greater than one.