The vertices of the hyperbola 9x2−16y2−36x+96y−252=0 are
A
and
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
and
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
and
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 B and We have, 9(x2−4x+4)−16(y2−6y+9)=144 ⇒(x−2)242−(y−3)232=1 Shifting the origin at (2, 3), we have x242−y232=1 Where, x = X + 2, y = Y + 3. Then coordinates of the vertices are x=±4,Y=3 i.e. x=6,y=3 and x=−2,y=3