The equation of the ellipse whose axes are coincident with the coordinates axes and which touches the straight lines3x−2y−20=0 and x+6y−20=0 is
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
x240+y210=1
b
x25+y28=1
c
x210+y240=1
d
x240+y230=1
answer is A.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Let the equation of the ellipse bex2a2+y2b2=1We know that the general equation of the tangent to the ellipse isy=mx±a2m2+b2----(i) Since 3x−2y−20=0 or y=32x−10 is tangent to the ellipse, comparing with (i), m=32 and a2m2+b2=100 or a2×94+b2=100 or 9a2+4b2=400-----(ii) Similarly, since x+6y−20=0 , i.e., y=−16x+103 is tangent to the ellipse, comparing with (i), m=16 and a2m2+b2=1009 or a236+b2=1009 or a2+36b2=400-----(iii) Solving (ii) and (iii), we get a2=40 and b2=10 . Therefore, the required equation of the ellipse is x240+y210=1