Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6

Q.

On the ellipse x28+y24=1.  let P be a point in the second quadrant such that the tangent at P to the ellipse is perpendicular to the line x + 2y = 0. Let S and S' be the foci of the ellipse and e be its eccentricity. If A is the area of the triangle SPS' then, the value of 5-e2 A 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

answer is 6.

(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

e=8-48=48=12 slope of Tangent m=-1-12=2 point P=-a2mc, b2c =-8×232+4, 432+4(c2=a2m2+b2) =-166, 46 area of SPS1 is =12×2ae×46 =8-4 × 46 =86=43=A (5-e2)A=5-12.43                =366=6

Watch 3-min video & get full concept clarity

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon