If ${e_1}$ and ${e_2}$ are respectively the eccentricities of the ellipse $\dfrac{{{x^2}}}{{18}} + \dfrac{{{y^2}}}{4} = 1$ and the hyperbola $\dfrac{{{x^2}}}{9} - \dfrac{{{y^2}}}{4} = 1$, then the relation between ${e_1}$ and ${e_2}$ is:$ {\text{A}}{\text{. }}3{\left( {{e_1}} \right)^2} + {\left( {{e_2}} \right)^2} = 2 $${\text{B}}{\text{. }}{\left( {{e_1}} \right)^2} + 2{\left( {{e_2}} \right)^2} = 3 $$ {\text{C}}{\text{. 2}}{\left( {{e_1}} \right)^2} + {\left( {{e_2}} \right)^2} = 3 $$ {\text{D}}{\text{. }}{\left( {{e_1}} \right)^2} + 3{\left( {{e_2}} \right)^2} = 2 $
Last updated date: 19th Mar 2023
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Answer
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Hint: Here, we will be using the formulas for the eccentricities of any general ellipse $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ where $a > b$ and any general hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$ which are ${e_1} = \sqrt {1 - \left( {\dfrac{{{b^2}}}{{{a^2}}}} \right)} {\text{ }}$ and ${e_2} = \sqrt {1 + \left( {\dfrac{{{b^2}}}{{{a^2}}}} \right)} $ respectively.
Complete step-by-step answer:
Complete step-by-step answer:
Given equation of ellipse is $\dfrac{{{x^2}}}{{18}} + \dfrac{{{y^2}}}{4} = 1 \Rightarrow \dfrac{{{x^2}}}{{{{\left( {3\sqrt 2 } \right)}^2}}} + \dfrac{{{y^2}}}{{{2^2}}} = 1{\text{ }} \to {\text{(1)}}$
and equation of hyperbola is $\dfrac{{{x^2}}}{9} - \dfrac{{{y^2}}}{4} = 1 \Rightarrow \dfrac{{{x^2}}}{{{3^2}}} - \dfrac{{{y^2}}}{{{2^2}}} = 1{\text{ }} \to {\text{(2)}}$
As we know the formula for eccentricity of any general ellipse $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1{\text{ }} \to {\text{(3)}}$ where $a > b$ is given by
${e_1} = \sqrt {1 - \left( {\dfrac{{{b^2}}}{{{a^2}}}} \right)} {\text{ }} \to {\text{(4)}}$
On comparing equations (1) and (3), we get
$a = 3\sqrt 2 $ and $b = 2$
Clearly, $ 3\sqrt 2 > 2$ i.e., $b > a$ so the formula given by equation (4) is valid in this case.
Putting the above values in equation (4), we get
${e_1} = \sqrt {1 - \left( {\dfrac{{{2^2}}}{{{{\left( {3\sqrt 2 } \right)}^2}}}} \right)} {\text{ = }}\sqrt {1 - \left( {\dfrac{4}{{18}}} \right)} = \sqrt {\dfrac{{18 - 4}}{{18}}} = \sqrt {\dfrac{{14}}{{18}}} = \sqrt {\dfrac{7}{9}}$
$\Rightarrow {\left( {{e_1}} \right)^2} = \dfrac{7}{9}{\text{ }} \to {\text{(5)}} $
Also we know the formula for eccentricity of any general hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1{\text{ }} \to {\text{(6)}}$ is given by
${e_2} = \sqrt {1 + \left( {\dfrac{{{b^2}}}{{{a^2}}}} \right)} {\text{ }} \to {\text{(7)}}$
On comparing equations (2) and (6), we get
$a = 3$ and $b = 2$
Putting the above values in equation (7), we get
${e_2} = \sqrt {1 + \left( {\dfrac{{{2^2}}}{{{3^2}}}} \right)} {\text{ = }}\sqrt {1 + \left( {\dfrac{4}{9}} \right)} = \sqrt {\dfrac{{9 + 4}}{9}} = \sqrt {\dfrac{{13}}{9}} $
$\Rightarrow {\left( {{e_2}} \right)^2} = \dfrac{{13}}{9}{\text{ }} \to {\text{(8)}} $
Now, we will use the equations (5) and (8) to verify which one of the relations given in the options is correct.
Now, taking LHS of option A i.e., $3{\left( {{e_1}} \right)^2} + {\left( {{e_2}} \right)^2} = 3 \times \left( {\dfrac{7}{9}} \right) + \dfrac{{13}}{9} = \dfrac{{21}}{9} + \dfrac{{13}}{9} = \dfrac{{34}}{9} \ne 2 = {\text{RHS}}$
Now, taking LHS of option B i.e., ${\left( {{e_1}} \right)^2} + 2{\left( {{e_2}} \right)^2} = \dfrac{7}{9} + 2 \times \left( {\dfrac{{13}}{9}} \right) = \dfrac{7}{9} + \dfrac{{26}}{9} = \dfrac{{33}}{9} = \dfrac{{11}}{3} \ne 3 = {\text{RHS}}$
Now, taking LHS of option C i.e., $2{\left( {{e_1}} \right)^2} + {\left( {{e_2}} \right)^2} = 2 \times \left( {\dfrac{7}{9}} \right) + \dfrac{{13}}{9} = \dfrac{{14}}{9} + \dfrac{{13}}{9} = \dfrac{{27}}{9} = 3 = {\text{RHS}}$
Now, taking LHS of option B i.e., ${\left( {{e_1}} \right)^2} + 3{\left( {{e_2}} \right)^2} = \dfrac{7}{9} + 3 \times \left( {\dfrac{{13}}{9}} \right) = \dfrac{7}{9} + \dfrac{{39}}{9} = \dfrac{{46}}{9} \ne 2 = {\text{RHS}}$
Clearly only option C is satisfied so the required relation is $2{\left( {{e_1}} \right)^2} + {\left( {{e_2}} \right)^2} = 3$.
Hence, option C is correct.
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