On the ellipse x^2/8 + y^2/4 =1 let P be a point in the second quadrant such that the tangent at P to the ellipse is perpendicular to the line

Question

On the ellipse \(\cfrac{x^2}{8}+\cfrac{y^2}{4} =1\) 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 :

(1) 6

(2) 12

(3) 14

(4) 24

Answer 1

Correct answer is: (1) 6

Equation of tangent : y = 2x + 6 

at P

e = 1/√2

S & S' = (–2, 0) & (2, 0)