Let the director circle of the hyperbola H:x2a2−y2b2=1 be denoted by C1 and let director circle of ellipse E:x2a2+y2b2=13 be denoted C2 . If the circle C2 is also the director circle of C1 and let e1 and e2 be the eccentricities of the ellipse E and hyperbola H, then e22e12 is equal to (a>b)

∴ C1:x2+y2=a2−b2⇒ C2:x2+y2=a2+b23Director circle of C1is x2+y2=2(a2−b2)⇒ 2(a2−b2)=a2+b23⇒ 6a2−6b2=a2+b2⇒ 5a2=7b25a2=7a2(1−e12)⇒e12=27 and 5a2=7a2(e22−1)⇒e12=127

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Let the director circle of the hyperbola $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ be denoted by $C_{1}$ and let director circle of ellipse $E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = \frac{1}{3}$ be denoted $C_{2}$ . If the circle $C_{2}$ is also the director circle of $C_{1}$ and let $e_{1} and e_{2}$ be the eccentricities of the ellipse E and hyperbola H, then $\frac{e_{2}^{2}}{e_{1}^{2}}$ is equal to $(a > b)$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 6.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$∴ C_{1} : x^{2} + y^{2} = a^{2} - b^{2} \Rightarrow C_{2} : x^{2} + y^{2} = \frac{a^{2} + b^{2}}{3}$

Director circle of $C_{1}$ is $x^{2} + y^{2} = 2 (a^{2} - b^{2}) \Rightarrow 2 (a^{2} - b^{2}) = \frac{a^{2} + b^{2}}{3}$

$\Rightarrow 6 a^{2} - 6 b^{2} = a^{2} + b^{2} \Rightarrow 5 a^{2} = 7 b^{2}$

$5 a^{2} = 7 a^{2} (1 - e_{1}^{2}) \Rightarrow e_{1}^{2} = \frac{2}{7} a n d 5 a^{2} = 7 a^{2} (e_{2}^{2} - 1) \Rightarrow e_{1}^{2} = \frac{12}{7}$

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!