Q.

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

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away
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)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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
Watch 3-min video & get full concept clarity
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon