JEE Main202228 Jul 2022Evening ShiftMathematicsEllipseActual
Let the tangents at the points P and Q on the ellipse x 2 2 + y 2 4 = 1 meet at the point R 2 , 2 2 - 2 . If S is the focus of the ellipse on its negative major axis, then S P 2 + S Q 2 is equal to
Correct answer
0
Step-by-step solution
Given ellipse is x 2 2 + y 2 4 = 1   . . . . . . . . . . . 1 , So its eccentricity will be a 2 = b 2 1 - e 2 ⇒ 2 = 4 1 - e 2 ⇒ 1 2 = 1 - e 2 ⇒ e = 1 2 So, focus S will be S ≡ 0 , - a e ≡ 0 , - 2 Now, equation of chord of contact will be T = 0 ⇒ x 2 + 2 2 - 2 y 4 = 1 ⇒ x 2 = 1 - 2 - 1 y 2   . . . . . . . . . . 2 Now on solving equation 1   &   2 we get, ⇒ y = 0 , 2    &   x = 2 , 1 So points P   &   Q is given by