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

All the chords of the hyperbola 3x2−y2−2x+4y=0, subtending a right angle at the origin pass through the fixed point :

A
(1,2)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
(1,2)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
(1,2)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
None of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A (1,2)
Let ax+by=1 be the chord ...(1)
Making the equation of hyperbola homogeneous using (1), we get 3x2y2+(2x+4y)(ax+by)=0
(32a)x2+(1+4b)y2+(2b+4a)xy=0
Since the angle subtended at the origin is a right angle,so, coefficient of x2+ coefficient of y2=0
(32a)+(1+4b)=0a=2b+1
The chords are (2b+1)x+by1=0
b(2+y)+(x1)=0, which clearly pass through the fixed point (1,2).

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