On the ellipse [x^2 8 + y^2 4 = 1] let P be a point in the second quadrant such that the t…

Question

On the ellipse $${{{x^2}} \over 8} + {{{y^2}} \over 4} = 1$$ 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 :

Accepted Answer

Equation of tangent : y = 2x + 6 at P

$$\therefore$$ P(-8/3, 2/3)

$$e = {1 \over {\sqrt 2 }}$$

S \& S' = (-2, 0) \& (2, 0)

Area of $$\Delta$$SPS' = $${1 \over 2} \times 4 \times {2 \over 3}$$

A = $${4 \over 3}$$

$$\therefore$$ (5 - e2)A = $$\left( {5 - {1 \over 2}} \right)$$ ${4 \over 3}$

Solutions · 1