A normal with slope 1 6 is drawn from the point (0,- ) to the parabola x2=-4ay, where a>0. Let L be the line passing through (0,- ) and parallel to the directrix of the parabola. Suppose that L intersects the parabola at two points A and B. Let r denote the length of the latus rectum and s denote the square of the length of the line segment AB. If r:s=1:16, then the value of 24a is .
Answer
12
Solution
Point on x2=-4ay: (2at,-at2); tangent slope dydx=- x2a=-t, normal slope 1t.
1t= 1 6 t= 6, foot (2 6a,-6a). Normal: y+6a= 1 6(x-2 6a); at x=0, y=-8a, so =8a.
L: y=-8a (parallel to directrix y=a): x2=32a2, AB=8 2a, s=128a2.
r=4a: 4a128a2= 116 a= 12, so 24a=12.
Answer: 12