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 :
Option: 1 (8,12)
Option: 2 \left ( \frac{20}{3},12 \right )
Option: 3 \left ( \frac{24}{5},10 \right )
Option: 4 \left ( 8,10 \right )

Answers (1)

\begin{aligned} &\text { For ellipse } \frac{x^{2}}{25}+\frac{y^{2}}{b^{2}}=1 \quad(b<5)\\ &\text { Let } e_{1} \text { is eccentricity of ellipse }\\ &\therefore \quad b^{2}=25\left(1-e_{1}^{2}\right) \ldots \ldots(1) \end{aligned}

For Hyperbola

\begin{aligned} &\frac{x^{2}}{16}-\frac{y^{2}}{b^{2}}=1\\ &\text { Let } \mathrm{e}_{2} \text { is eccentricity of hyperbola. }\\ &\therefore \quad \mathrm{b}^{2}=16\left(\mathrm{e}_{2}^{2}-1\right) \quad \ldots \ldots(2) \end{aligned}

From (1) and (2)

\begin{aligned} &25\left(1-\mathrm{e}_{1}^{2}\right)=16\left(\mathrm{e}_{2}^{2}-1\right)\\ &\text { Now } \mathrm{e}_{1} \cdot \mathrm{e}_{2}=1 \quad \text { (given) }\\ &\therefore \quad 25\left(1-\mathrm{e}_{1}^{2}\right)=16\left(\frac{1-\mathrm{e}_{1}^{2}}{\mathrm{e}_{1}^{2}}\right)\\ &\text { or }\\ &e_{1}=\frac{4}{5} \quad \therefore e_{2}=\frac{5}{4} \end{aligned}

distance beween foci = 2ae

\begin{aligned} &\therefore \text { distance for ellipse }=2 \times 5 \times \frac{4}{5}=8=\alpha\\ &\text { distance for hyperbola }=2 \times 4 \times \frac{5}{4}=10=\beta\\ &\therefore(\alpha, \beta) \equiv(8,10) \end{aligned}

 

Most Viewed Questions

Preparation Products

Knockout JEE Main April 2021 (One Month)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 14000/- ₹ 4999/-
Buy Now
Knockout JEE Main May 2021

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 22999/- ₹ 9999/-
Buy Now
Test Series JEE Main May 2021

Unlimited Chapter Wise Tests, Unlimited Subject Wise Tests, Unlimited Full Mock Tests, Get Personalized Performance Analysis Report,.

₹ 6999/- ₹ 2999/-
Buy Now
Knockout JEE Main May 2022

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 34999/- ₹ 14999/-
Buy Now
JEE Main Rank Booster 2021

Booster and Kadha Video Lectures, Unlimited Full Mock Test, Adaptive Time Table, 24x7 Doubt Chat Support,.

₹ 13999/- ₹ 6999/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
 
Exams
Articles
Questions