Let e1 and e2 be the eccentricities of the ellipse,


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 :

  1. (1) $(8,12)$

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

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

  4. (4) $(8,10)$

Correct Option: , 4


Equation of ellipse is $\frac{x^{2}}{25}+\frac{y^{2}}{b^{2}}=1$

Then, $e_{1}=\sqrt{1-\frac{b^{2}}{25}}$

The equation of hyperbola, $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$

Then, $e_{2}=\sqrt{1+\frac{b^{2}}{16}}$

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

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

$\Rightarrow 1+\frac{b^{2}}{16}-\frac{b^{2}}{25}-\frac{b^{4}}{25 \times 16}=1$

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

$\therefore e_{1}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}$

And, $e_{2}=\sqrt{1+\frac{9}{16}}=\frac{5}{4}$

Distance between focii of ellipse

$=\alpha=2 a e_{1}=2(5)\left(e_{1}\right)=8$

Distance between focii of hyperbola

$=\beta=2 a e_{2}=2(4)\left(e_{2}\right)=10$

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

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now