Let P be an extremity of the latus rectum of an ellipse E: x^2 a^2 + y^2 b^2 = 1 ( a > b ) in the first quadrant. The normal to E at P meets the x -axis at G and the y -axis at K . If the area of the triangle OGK (where O is the origin) is equal to the area of the triangle formed by the origin and the endpoints of the corresponding latus rectum, then the square of the eccentricity of the ellipse is equal to

Step-by-step solution

The coordinates of the extremity of the latus rectum in the first quadrant are P (ae, b^2 a ) . The equation of the normal to the ellipse at a point (x₁, y₁) is a^2 x x₁ - b^2 y y₁ = a^2 e^2 . Substituting the coordinates of P , the normal is: a^2 x ae - b^2 y b^2 a = a^2 e^2 ax e - ay = a^2 e^2 To find the x -intercept G , put y = 0 : ax e = a^2 e^2 x = ae^3 . Thus, G (ae^3, 0) . To find the y -intercept K , put x = 0 : -ay = a^2 e^2 y = -ae^2 . Thus, K (0, -ae^2) . The area of OGK is: 1 2 |ae^3| |-ae^2| = 1 2 a^2