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 $
Answer
361.8k+ views
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.
Last updated date: 28th Sep 2023
•
Total views: 361.8k
•
Views today: 7.61k
Recently Updated Pages
What is the Full Form of DNA and RNA

What are the Difference Between Acute and Chronic Disease

Difference Between Communicable and Non-Communicable

What is Nutrition Explain Diff Type of Nutrition ?

What is the Function of Digestive Enzymes

What is the Full Form of 1.DPT 2.DDT 3.BCG

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Drive an expression for the electric field due to an class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

Derive an expression for electric potential at point class 12 physics CBSE

Why is the electric field perpendicular to the equipotential class 12 physics CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE
