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

The eccentricity of the hyperbola in the standard form x2a2+y2b2=1, passing through 3,0 and 32,2 is


A

133

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

13

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

3

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

133

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

53

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

The correct option is D

133


Explanation for the correct answer:

Finding the eccentricity:

The standard form of a hyperbola is x2a2+y2b2=1.

The hyperbola passes through 3,0.

So, put x=3 and y=0 in x2a2+y2b2=1

32a2+0=1a2=9

The hyperbola passes through 32,2.

So, put x=32 , y=2 and a2=9 in x2a2-y2b2=1

3229-22b2=1189-4b2=12-4b2=14b2=1b2=4

The eccentricity of the hyperbola is given by e=1+b2a2.

Put a2=9,b2=4.

e=1+49=139=133

Hence, the correct answer is option (D)


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Hyperbola and Terminologies
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon