Consider the parabola P : y2=4kx and the ellipse E : x2a2+ y2b2=1. Let the line segment joining the points of intersection of P and E, be their latus rectums. If the eccentricity of E is e, then e2+2 2 is equal to .
Answer
3
Solution
The ends of the latus rectum of P are (k, 2k) and those of E are (ae, b2a ). Since they coincide, k = ae and 2k = b2a, so
2ae = a2(1 - e2)a = a(1 - e2)
e2 + 2e - 1 = 0 e = 2 - 1 (∵ 0<e<1)
e2 = 3 - 2 2
∴ e2 + 2 2 = 3
Answer: 3