
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
162.6k+ 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.
Recently Updated Pages
Fluid Pressure - Important Concepts and Tips for JEE

JEE Main 2023 (February 1st Shift 2) Physics Question Paper with Answer Key

Impulse Momentum Theorem Important Concepts and Tips for JEE

Graphical Methods of Vector Addition - Important Concepts for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 1) Physics Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NEET 2025 – Every New Update You Need to Know

Verb Forms Guide: V1, V2, V3, V4, V5 Explained

NEET Total Marks 2025

1 Billion in Rupees
