For what value of k is the parabola y=k-x^2 viewed at a right angle * from point A(0,1)?


* i.e. displaying a right angle between the two tangents.

To pass through (0,1) the line must have an equation of the form
y= mx+1
Substitute for y = k-x^2

mx+1=k-x^2
x^2+mx+1-k=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*(1-k)=0
m=ħsqrt(4(1-k))

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(1-k))=-1
1-k = 1/4
k=3/4

Posted by Jer on 2011-10-28 13:42:03