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
Ends of the latus rectum of P: (k, 2k); of E: (ae, b2a )
They coincide k=ae and 2k= b2a
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