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 x + 2 y = 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 - e 2 · A is

Step-by-step solution

We have an ellipse x 2 8 + y 2 4 = 1 , we have a general point P ( 8 cos θ , 2 sin θ ) Equation of tangent at point p x 8 cos θ + y 2 sin θ - 1 = 0 Slope = - 1 2 cot θ Given that, tangent at P is perpendicular to the line x + 2 y = 0 . So, product of their slopes are perpendicular ⇒ - 1 2 cot θ × - 1 2 = - 1 ⇒ - 1 2 cot θ = 2 ⇒ cot θ = - 2 2 ⇒ cos θ = - 2 2 3 , sin θ = 1 3 So, point P - 8 3 , 2 3 A = 1 2 × 2 ae × 2 3 Where, e =