Let $e_{1}$ and $e_{2}$ be the eccentricities of the ellipse, $\frac{x^{2}}{25}+\frac{y^{2}}{b^{2}}=1(b<5)$ and the hyperbola, $\frac{x^{2}}{16}-\frac{y^{2}}{b^{2}}=1$ respectively satisfying $e_{1} e_{2}=1$. 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 :

(B) $\left(\frac{24}{5}, 10\right)$

(C) $\left(\frac{20}{3}, 12\right)$

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Hyperbola

Let

e_{1}

and

e_{2}

be the eccentricities of the ellipse,

\frac{x ^{2}}{25} + \frac{y ^{2}}{b ^{2}} = 1 (b < 5)

and the hyperbola,

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

respectively satisfying

e_{1} e_{2} = 1

. If

α

and

β

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

(α, β)

is equal to :

Answer: (A)

For the ellipse

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

e_{1} = 1 - \frac{b ^{2}}{25}

, and for the hyperbola

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

e_{2} = 1 + \frac{b ^{2}}{16}

. The condition

e_{1} e_{2} = 1

squares to

(1 - \frac{b ^{2}}{25}) (1 + \frac{b ^{2}}{16}) = 1

, which expands to

\frac{9 b ^{2}}{400} - \frac{b ^{4}}{400} = 0

, giving

b^{2} = 9

. Then

e_{1} = \frac{4}{5}

and

e_{2} = \frac{5}{4}

. The distance between the ellipse's foci is

α = 2 \times 5 \times \frac{4}{5} = 8

, and between the hyperbola's foci is

β = 2 \times 4 \times \frac{5}{4} = 10

, so

(α, β) = (8, 10)

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Single correct option