Solveeit Logo

Question

Question: If \({e_1}\) and \({e_2}\) are respectively the eccentricities of the ellipse \(\dfrac{{{x^2}}}{{18}...

If e1{e_1} and e2{e_2} are respectively the eccentricities of the ellipse x218+y24=1\dfrac{{{x^2}}}{{18}} + \dfrac{{{y^2}}}{4} = 1 and the hyperbola x29y24=1\dfrac{{{x^2}}}{9} - \dfrac{{{y^2}}}{4} = 1, then the relation between e1{e_1} and e2{e_2} is:

A. 3(e1)2+(e2)2=2{\text{A}}{\text{. }}3{\left( {{e_1}} \right)^2} + {\left( {{e_2}} \right)^2} = 2

B. (e1)2+2(e2)2=3{\text{B}}{\text{. }}{\left( {{e_1}} \right)^2} + 2{\left( {{e_2}} \right)^2} = 3

C. 2(e1)2+(e2)2=3{\text{C}}{\text{. 2}}{\left( {{e_1}} \right)^2} + {\left( {{e_2}} \right)^2} = 3

D. (e1)2+3(e2)2=2{\text{D}}{\text{. }}{\left( {{e_1}} \right)^2} + 3{\left( {{e_2}} \right)^2} = 2

Explanation

Solution

Here, we will be using the formulas for the eccentricities of any general ellipse x2a2+y2b2=1\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1 where a>ba > b and any general hyperbola x2a2y2b2=1\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1 which are e1=1(b2a2) {e_1} = \sqrt {1 - \left( {\dfrac{{{b^2}}}{{{a^2}}}} \right)} {\text{ }} and e2=1+(b2a2){e_2} = \sqrt {1 + \left( {\dfrac{{{b^2}}}{{{a^2}}}} \right)} respectively.

Complete step-by-step answer:

Given equation of ellipse is x218+y24=1x2(32)2+y222=1 (1)\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 x29y24=1x232y222=1 (2)\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 x2a2+y2b2=1 (3)\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1{\text{ }} \to {\text{(3)}} where a>ba > b is given by

e1=1(b2a2) (4){e_1} = \sqrt {1 - \left( {\dfrac{{{b^2}}}{{{a^2}}}} \right)} {\text{ }} \to {\text{(4)}}

On comparing equations (1) and (3), we get

a=32a = 3\sqrt 2 and b=2b = 2

Clearly, 32>2 3\sqrt 2 > 2 i.e., b>ab > a so the formula given by equation (4) is valid in this case.

Putting the above values in equation (4), we get

e1=1(22(32)2) = 1(418)=18418=1418=79{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}}

(e1)2=79 (5)\Rightarrow {\left( {{e_1}} \right)^2} = \dfrac{7}{9}{\text{ }} \to {\text{(5)}}

Also we know the formula for eccentricity of any general hyperbola x2a2y2b2=1 (6)\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1{\text{ }} \to {\text{(6)}} is given by

e2=1+(b2a2) (7){e_2} = \sqrt {1 + \left( {\dfrac{{{b^2}}}{{{a^2}}}} \right)} {\text{ }} \to {\text{(7)}}

On comparing equations (2) and (6), we get

a=3a = 3 and b=2b = 2

Putting the above values in equation (7), we get

e2=1+(2232) = 1+(49)=9+49=139{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}}

(e2)2=139 (8)\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(e1)2+(e2)2=3×(79)+139=219+139=3492=RHS3{\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., (e1)2+2(e2)2=79+2×(139)=79+269=339=1133=RHS{\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(e1)2+(e2)2=2×(79)+139=149+139=279=3=RHS2{\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., (e1)2+3(e2)2=79+3×(139)=79+399=4692=RHS{\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(e1)2+(e2)2=32{\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.