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 which of the following relation between ${e_1}$ and ${e_2}$ is correct?
(A) $3e_1^2 + e_2^2 = 2$
(B) $e_1^2 + 2e_2^2 = 3$
(C) $2e_1^2 + e_2^2 = 3$
(D) $e_1^2 + 3e_2^2 = 2$

Answer 1

As we know that the general equation of ellipse $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1{\text{ or }}\dfrac{{{y^2}}}{{{a^2}}} + \dfrac{{{x^2}}}{{{b^2}}} = 1{\text{ where }}a > b$ will have an eccentricity of $\sqrt {1 - \dfrac{{{b^2}}}{{{a^2}}}}$ and general hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1{\text{ or }}\dfrac{{{y^2}}}{{{a^2}}} - \dfrac{{{x^2}}}{{{b^2}}} = 1$ will have an eccentricity of $\sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}}$. Use this information to find the eccentricities ${e_1}{\text{ and }}{e_2}$ and check for the correct option.

Complete step by step solution:
In this problem, we are given with equations of two curves, i.e. an ellipse $\dfrac{{{x^2}}}{{18}} + \dfrac{{{y^2}}}{4} = 1$ and a hyperbola with an equation $\dfrac{{{x^2}}}{9} - \dfrac{{{y^2}}}{4} = 1$ . And we need to find a relation between the eccentricities of these two curves.
Before starting with the solution we should know a few things about eccentricity and how to calculate it. In mathematics, the eccentricity of a conic section is a non-negative real number that uniquely characterizes its shape. One can think of the eccentricity as a measure of how much a conic section deviates from being circular. Any conic section can be defined as the locus of points whose distances to a point (the focus) and a line (the directrix) are in a constant ratio. That ratio is called the eccentricity, commonly denoted as $'e'$ .
The eccentricity of a circle is zero. For an ellipse which is not a circle is greater than zero but less than one. The eccentricity of a parabola is $1$ and for a hyperbola, it is greater than $1$ .
For an ellipse with an equation $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1{\text{ or }}\dfrac{{{y^2}}}{{{a^2}}} + \dfrac{{{x^2}}}{{{b^2}}} = 1{\text{ where }}a > b$ , the eccentricity $\left( e \right)$ will be given by $\sqrt {1 - \dfrac{{{b^2}}}{{{a^2}}}}$ .
For a hyperbola with an equation $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1{\text{ or }}\dfrac{{{y^2}}}{{{a^2}}} - \dfrac{{{x^2}}}{{{b^2}}} = 1$ , the eccentricity $\left( e \right)$ will be given by $\sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}}$ .
So, for the given ellipse $\dfrac{{{x^2}}}{{18}} + \dfrac{{{y^2}}}{4} = 1$ , we have ${a^2} = 18{\text{ and }}{b^2} = 4$ the eccentricity $\left( {{e_1}} \right)$ will be given by:
$\Rightarrow {e_1} = \sqrt {1 - \dfrac{{{b^2}}}{{{a^2}}}} = \sqrt {1 - \dfrac{4}{{18}}} = \sqrt {\dfrac{{14}}{{18}}}$
Therefore, on further simplifying the value, we get: ${e_1} = \sqrt {\dfrac{7}{9}}$
Also, for the given hyperbola $\dfrac{{{x^2}}}{9} - \dfrac{{{y^2}}}{4} = 1$ , we have ${a^2} = 9{\text{ and }}{b^2} = 4$ the eccentricity $\left( {{e_2}} \right)$ will be given by:
$\Rightarrow {e_2} = \sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}} = \sqrt {1 + \dfrac{4}{9}} = \sqrt {\dfrac{{13}}{9}}$
Therefore, we get: ${e_2} = \sqrt {\dfrac{{13}}{9}}$
Since, we have to find the correct relationship among the given option, let’s check the squares of these values of eccentricities:
$\Rightarrow e_1^2 = {\left( {\sqrt {\dfrac{7}{9}} } \right)^2} = \dfrac{7}{9}{\text{ and }}e_2^2 = {\left( {\sqrt {\dfrac{{13}}{9}} } \right)^2} = \dfrac{{13}}{9}$
As we know the options are given in the form of equations with $e_1^2{\text{ and }}e_2^2$. So now we can check the left side of the options to find the correct answer:
For option (A), $3e_1^2 + e_2^2 = 3 \times \dfrac{7}{9} + \dfrac{{13}}{9} = \dfrac{{21 + 13}}{9} = \dfrac{{44}}{9} \ne 2$
For option (B), $e_1^2 + 2e_2^2 = \dfrac{7}{9} + 2 \times \dfrac{{13}}{9} = \dfrac{{7 + 26}}{9} = \dfrac{{33}}{9} \ne 3$
For option (C), $2e_1^2 + e_2^2 = 2 \times \dfrac{7}{9} + \dfrac{{13}}{9} = \dfrac{{14 + 13}}{9} = \dfrac{{27}}{9} = 3$
For option (D), $e_1^2 + 3e_2^2 = \dfrac{7}{9} + 3 \times \dfrac{{13}}{9} = \dfrac{{7 + 39}}{9} = \dfrac{{46}}{9} \ne 2$

Hence, the option (C) is the correct answer.

Note:
The use of a general form of equation played a crucial part in the solution. Notice that we compared the given equation of curves with their general form to find the values of ${a^2}$ and ${b^2}$ for both ellipse and hyperbola. Be careful while putting the values in the expression of eccentricities.