Question 935929
{{{-2y=-x+3}}}
{{{y=x/2-3/2}}}
You want a slope of the desired line to be {{{-2}}}, negative reciprocal of that slope.  You do not know the y-intercept of {{{y=-2x+k}}}.  You want THIS line to intersect the parabola AT ONE POINT, ....


{{{x^2=16(-2x+k)}}}
{{{x^2=-32x+16k}}}
{{{x^2+32x-16k=0}}}, you need discriminant of this to be 0.  This will ensure the intersection will be ONE POINT.


{{{(-32)^2-4*(-16k)=0}}}
{{{1024+64k=0}}}
{{{64k=-1024}}}
{{{highlight(k=-16)}}}


Now you have the line {{{y=-2x-16}}} and you want to know the intersection of this with the parabola {{{x^2=16y}}}.
-
{{{x^2=16(-2x-16)}}}
{{{x^2=-32x-256}}}
{{{x^2+32x+256=0}}}
{{{(x+16)^2=0}}}
THIS means, {{{highlight(x=-16)}}} and the corresponding coordinate y is {{{y=-2*(-16)-16}}}
{{{y=32-16}}}
{{{highlight(y=16)}}}


NEXT:
What is the line with slope {{{-2}}} and with the point (-16,16) ?
Use the point-slope form equation for a line to form the equation of this line.