Mathematics - Hyperbola Question with Solution | TestHub

MathematicsHyperbolaMiscellaneous/MixedMedium2 minPYQ_2020

MathematicsMediumsingle choice

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:

Options:

Answer:

Solution:

$e_{1} = \sqrt{1 - \frac{b^{2}}{25}}$ and $e_{2} = \sqrt{1 + \frac{b^{2}}{16}}$

Given $e_{1} e_{2} = 1$

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

$\Rightarrow \frac{b^{2}}{16} - \frac{b^{2}}{25} - \frac{b^{4}}{25 \times 16} = 0$

$\Rightarrow \frac{9}{16 \times 25} b^{2} - \frac{b^{4}}{25 \times 16} = 0 \Rightarrow b^{2} = 9$

Thus, $e_{1} = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$ and $e_{2} = \sqrt{1 + \frac{9}{16}} = \frac{5}{4}$

Therefore, $α = 2 (5) (e_{1}) = 8$ and $β = 2 (4) (e_{2}) = 10$

$∴ (α, β) = (8, 10)$

Stream:JEESubject:MathematicsTopic:HyperbolaSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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