Originally Posted by
skeeter maybe the question wants two distinct lines tangent to the curve that are perpendicular ... ?
if so, then given the symmetry of $\displaystyle f(x)$, one point of tangency would be $\displaystyle (t,f(t))$ and the other would be $\displaystyle (-t,f(-t))$. note also that since $\displaystyle f(x)$ is even, $\displaystyle f(t) = f(-t)$
$\displaystyle f(x) = kx^2+3$
$\displaystyle f'(x) = 2kx$
$\displaystyle f'(t) = 2kt$
$\displaystyle f'(-t) = -2kt$
$\displaystyle f'(t)$ and $\displaystyle f'(-t)$ are opposite reciprocals ...
$\displaystyle 2kt = -\frac{1}{-2kt}$
$\displaystyle 4k^2t^2 = 1$
$\displaystyle t = \frac{1}{|2k|}$
slopes ...
at $\displaystyle (t, f(t))$ , $\displaystyle m = 2k \cdot \frac{1}{|2k|} = 1$ if $\displaystyle k > 0$ and $\displaystyle -1$ if $\displaystyle k < 0$
at $\displaystyle (-t,f(t))$ , $\displaystyle m = -2k \cdot \frac{1}{|2k|} = -1$ if $\displaystyle k > 0$ and $\displaystyle 1$ if $\displaystyle k < 0$
it should be obvious that the two tangent lines intersect on the y-axis
one of the tangent line equations is ...
$\displaystyle y - (kt^2+3) = x - t$
for $\displaystyle x = 0$ ...
$\displaystyle y - kt^2 - 3 = -t$
$\displaystyle y = kt^2 - t + 3$
$\displaystyle y = k \cdot \frac{1}{4k^2} - \frac{1}{|2k|} + 3$
$\displaystyle y = \frac{1}{4k} - \frac{1}{|2k|} + 3$
the intersection point is $\displaystyle \left(0,\frac{1}{4k} - \frac{1}{|2k|} + 3\right)$
really ... a very poorly worded question.