
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
161.1k+ views
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.
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.
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