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10.2 parabola,ellipse,hyperbola — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHS10.2 parabola,ellipse,hyperbola1 Mark
MCQ
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
  • ✓
    $(8,10)$
  • B
    $(8,12)$
  • C
    $\left(\frac{20}{3}, 12\right)$
  • D
    $\left(\frac{24}{5}, 10\right)$
✓

Answer

Correct option: A.
$(8,10)$
a
For ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{b^{2}}=1 \quad(b<5)$

Let e $_{1}$ is eccentricity of ellipse

$\therefore \quad b^{2}=25\left(1-e_{1}^{2}\right) \ldots \ldots .(1)$

Again for hyperbola

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

Let $e _{2}$ is eccentricity of hyperbola.

$\therefore \quad b^{2}=16\left(e_{2}^{2}-1\right) \quad \ldots \ldots$

by (1)$\&(2)$

$25\left(1- e _{1}^{2}\right)=16\left( e _{2}^{2}-1\right)$

Now $e _{1} e _{2}=1 \quad$ (given)

$\therefore \quad 25\left(1- e _{1}^{2}\right)=16\left(\frac{1- e _{1}^{2}}{ e _{1}^{2}}\right)$

or $\quad e _{1}=\frac{4}{5} \quad \therefore e _{2}=\frac{5}{4}$

Now distance between foci is $2 ae$

$\therefore$ distance for ellipse $=2 \times 5 \times \frac{4}{5}=8=\alpha$

distance for hyperbola $=2 \times 4 \times \frac{5}{4}=10=\beta$

$\therefore(\alpha, \beta) \equiv(8,10)$

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