Question 200956
Have you had calculus? That's the
easiest way to do it. If the equation 
is {{{y = 16x^2}}} , then the equation 
of the slope at every point is
{{{y' = 32x}}}.
If I say {{{32x = 1.75}}} and
{{{32x = -1.75}}}, These are the x-
values of the points of contact
{{{32x = 1.75}}}
{{{x = .0546875}}}
and
{{{32x = -1.75}}}
{{{x = -.0546875}}}
And the y-coordinates are:
{{{y = 16x^2}}}
{{{y = 16*(.0546875)^2}}}
{{{y = 16*2.9907226}}}
{{{y = .0478515}}}
The y-coodinate for the other value is
the same, since parabola is symmetric
about y-axis
P(-.05469, .04785) and
Q(.05469, .04785) are the points of contact in mm