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 $

Question

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 $

Vedantu Content Team · Accepted Answer

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: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.Note: In these types of problems using the general equation for any ellipse or hyperbola, values of various parameters used in the formulas for eccentricities of the ellipse and hyperbola are evaluated. After that with the help of options given, the relation which satisfies the evaluated values of the eccentricities is the required relation between the eccentricities of the given ellipse and hyperbola.

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