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

# All the chords of the hyperbola 3x2âˆ’y2âˆ’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! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
C
(1,2)
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
D
None of these
No worries! Weâ€˜ve got your back. Try BYJUâ€˜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 3x2−y2+(−2x+4y)(ax+by)=0(3−2a)x2+(−1+4b)y2+(−2b+4a)xy=0Since the angle subtended at the origin is a right angle,so, coefficient of x2+ coefficient of y2=0⇒(3−2a)+(−1+4b)=0⇒a=2b+1∴ The chords are (2b+1)x+by−1=0⇒b(2+y)+(x−1)=0, which clearly pass through the fixed point (1,−2).

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Chords and Pair of Tangents
MATHEMATICS
Watch in App
Join BYJU'S Learning Program