CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

For the hyperbola x2cos2αy2sin2α=1, when α changes, remains constant.


A
abscissa of vertices
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
abscissa of foci
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
eccentricity
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
directrix
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B abscissa of foci

Given hyperbola is,

x2cos2αy2sin2α=1

Comparing with a standard hyperbola we get,

a2=cos2α

b2=sin2α

We are asked to find which of the given options remain unchanged with change in α.

For finding that, we need to see which of the quantities remain independent of α.

b2=a2(e21)

e21=b2a2

=sin2cos2=tan2

e=|secα|

directrix,x=±ae=±cosαsecα

Abscissa of focus =±ae

=±cosα|secα|

=±1 which is independent of α


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Jumping to Details
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon