QJEEM20Mac48c: JEE Main 2020 Mathematics - Class 11 JEEM20Mac48c Solution: Let e 1 and e 2 be the eccentricities of the ellipse, (b < 5)…

Question

Verified Solution • 2025–26 Session moderate

Let e₁ and e₂ be the eccentricities of the ellipse,

{{{x^2}} \over {25}} + {{{y^2}} \over {{b^2}}} = 1

(b < 5) and the hyperbola,

{{{x^2}} \over {16}} - {{{y^2}} \over {{b^2}}} = 1

respectively satisfying e₁e₂ = 1. If

\alpha

and

\beta

are the distances between the foci of the
ellipse and the foci of the hyperbola
respectively, then the ordered pair (

\alpha

,

\beta

) is equal to :

Accepted Answer

For ellipse

{{{x^2}} \over {25}} + {{{y^2}} \over {{b^2}}} = 1\,\,\,(b &lt; 5)

Let e₁ is eccentricity of ellipse

\therefore

b² = 25 (1

-

{e_1}^2

) ........(1)

Again for hyperbola

{{{x^2}} \over {16}} - {{{y^2}} \over {{b^2}}} = 1

Let e₂ is eccentricity of hyperbola.

\therefore

{b^2} = 16({e_1}^2 - 1)

......(2)

by (1) & (2)

25(1 - {e_1}^2)\, = \,16({e_1}^2 - 1)

Now e₁ . e₂ = 1 (given)

\therefore

25(1 - {e_1}^2)\, = \,16\left( {{{1 - {e_1}^2} \over {{e_1}^2}}} \right)

or e₁ =

{4 \over 5}

\therefore

e₂ =

{5 \over 4}

Now distance between foci is 2ae

\therefore

Distance for ellipse =

2 \times 5 \times {4 \over 5} = 8 = \alpha

Distance for hyperbola =

2 \times 4 \times {5 \over 4} = 10 = \beta

\therefore

(

\alpha

,

\beta

)

\equiv

(8, 10)

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