The locus of the foot of perpendicular drawn from the centre of the ellipse x2+3y2=6 on any tangent to it is:
A
(x2−y2)2=6x2+2y2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(x2−y2)2=6x2−2y2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
(x2+y2)2=6x2+2y2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
(x2+y2)2=6x2−2y2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C(x2+y2)2=6x2+2y2 Let the foot of perpendicular drawn from the centre of ellipse on the tangent ′l′ is (h,k)
Slope of OP=kh ∴ Slope of tangent, m=−hk As we know, Equation of tangent is y=mx±√a2m2+b2...(1) put value of m in equation (1) and pass through (h,k) Thus, k=−hk⋅h±√6(−hk)2+2 ⇒k2+h2=±√6h2+2k2 ⇒(k2+h2)2=6h2+2k2 Put (x,y) at the place of (h,k) Hence, (x2+y2)2=6x2+2y2