** AP Calculus Question

"xo" = x subscript knot

Let f(x)=kx^2+c

a) Find xo in terms of k such that the tangent lines to the graph of f at (xo, f(xo)) and (-xo, f(-xo)) are perpendicular.

b) Find the slopes of the tangent lines mentioned in part a)

c) Find the coordinates, in terms of k and c, of the point of intersection of the tangent lines mentioned in part a)

I'm gonna let a = xo to make life easier with these text editors. (Also, I think you mean "naught" not "knot.")

a)
f '(x) = 2kx
f '(a) = 2ka
f '(-a) = -2ka
Since the tangent lines are perpendicular,
f '(a) = -1/(f '(-a))
2ka = -1/(-2ka)
2ka = 1/(2ka)
4k^2a^2 = 1
a^2 = 1/(4k^2)
a = 1/(2k) or a = -1/(2k)

b) slope of line with a = 1/(2k) is f '(1/(2k)) = 2k(1/(2k)) = 1
slope of line with a = -1/(2k) is f '(-1/(2k)) = 2k(-1/(2k)) = -1

c) For a = 1/(2k),
f(1/(2k)) = k(1/(2k))^2 + c = 1/(4k) + c
The line through (a, f(a)) with slope of 1 is
y - (1/(4k) + c) = 1(x - 1/(2k))
y - 1/(4k) - c = x - 1/(2k)
y = x - 1/(2k) + 1/(4k) + c
y = x - 1/(4k) + c
For a = -1/(2k),
f(-1/(2k)) = k(-1/(2k))^2 + c = 1/(4k) + c
The line's through (a, f(a)) with slope of -1 is
y - (1/(4k) + c) = -1(x - -1/(2k))
y - 1/(4k) - c = -x - 1/(2k)
y = -x - 1/(4k) + c
Tangent lines' point of intersection:
y = x - 1/(4k) + c
y = -x - 1/(4k) + c
--------------------------
2y = -2/(4k) + 2c
y = -1/(4k) + c
-1/(4k) + c = x - 1/(4k) + c
x = 0

f(x) = kx^2 + c

f '(x) = 2kx

(a.) 2k(-xo) = -1/(2kxo)

-4k²(xo)² = -1

xo = +/- 1/(2k)

(b.) The tangent line at (xo, f(xo)) has a slope of +/- 1 while the tangent line at (-xo, f(-xo)) also has a slope of +/- 1.

(c.) Find the equations of the tangents and set them both equal to each other to find the POI or the point of intersection.