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

If 2p2−3q2+4pq−p=0 and a variable line px+qy=1 always touches a parabola whose axis is parallel to X-axis, then equation of the parabola is

A
(y4)2=24(x2)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(y3)2=12(x1)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
(y4)2=12(x2)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
(y2)2=24(x4)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C (y4)2=12(x2)
The parabola be (ya)2=4b(xc)
Equation of tangent is (ya)=pq(xc)bqp
Comparing with px+qy=1, we get cp2bq2+apqp=0
c2=b3=a4=1 the equation is (y4)2=12(x2).

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Defining Conics
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon