wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The equation of the ellipse whose foci are at ±2,0 and eccentricity is 12, is x2a2+y2b2=1 Then what is the value of a2,b2.


A

a2=16,b2=12

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

a2=12,b2=16

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

a2=16,b2=4

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

a2=4,b2=16

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A

a2=16,b2=12


Explanation of correct option:

Finding values of a2,b2:

As we know, the foci of the ellipse having equation x2a2+y2b2=1 is ±ae,0(1).

Given, eccentricity e=12

Given, foci of parabola ae=±2,0

Equating points of parabola with equation 1.

2=ae2=a×12a=4

As we know, e2=1-b2a2

Substituting the values we get,

122=1-b24214=1-b216b216=34b2=12

Thus, the values are a2=16,b2=12.

Hence, option(A) is the correct answer.


flag
Suggest Corrections
thumbs-up
3
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Defining Conics
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon