F(x)=k(x^2)+3
= kx^2 + 3
F'(x) = 2kx
at( t, f(t)) slope of tangent is 2kt
at (-t, f(-t), slope of tangent is -2kt
but they are perpendicular, so
(2kt)(-2kt) = -1
4k^2t^2= 1
k^2 t^2 = 1/4
kt = ± 1/2
t = ± 1/ (2k)
when t = 1/(2k)
so at( t, f(t)) slope of tangent is 2kt = 2k(1/(2k)) = 1
at (-t, f(-t), slope of tangent is -2kt = -2k(1/(2k)) = -1
c) how about you try it ?
Consider the function F(x)=k(x^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 term of k.
b) Find the slopes of tangent lines mentioned in part (a).
c) Find the coordinates of the point of intersection of the tangent lines mentioned in part (a).
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