To pass through (0,1) the line must have an equation of the form
y= mx+1
Substitute for y = kx^2
mx+1=kx^2
x^2+mx+1k=0
In general a line intersects a parabola in two points. It will be tangent if there is only one solution to this equation. Hence the discriminant must equal zero:
m^2  4*1*(1k)=0
m=ħsqrt(4(1k))
So these are the slopes of the two tangents from the point (0,1). They will meet at a right angle if their product is 1, so we have
(4(1k))=1
1k = 1/4
k=3/4

Posted by Jer
on 20111028 13:42:03 