Start with the given parabola:
f(x)=kx²+3
The derivative is
f'(x)=2kx
(a)
We are given that the tangent lines at
(t, f(t)) and (-t,f(-t)) are perpendicular, which means that the product of the slopes =-1, or
f(t)*f(-t)=-1
2kt*2k(-t)=-1
Solving for t, we get
t=1/(2k) ....(1)
(b)
The slope of each tangent is given by
m1=f'(t)=2k/(2k)=1, ...(2) and
m2=f'(-t)=2k/(-2k)=-1 ...(3)
f(t)=k(t²)+3=1/(4k) + 3 ...(4)
f(-t)=k(-t)²+3=1/(4k) + 3 ...(5)
The equations of the tangent lines are therefore:
(y-f(t))=m1(x-t) ...(6)
(y-f(-t))=m2(x-(-t)) ...(7)
Solve (6) and (7) for x and y.
Hint: x=0 by symmetry.
f(x)=kx^2+3
a) if the tangent lines to the graph of f at (t,f(t)) and (-t, f(-t)) are perpendicular, find t in terms of k
b) Find the coordinate of the point of intersection of the tangent lines mentioned in part (a) in terms of k
1 answer
1 answer