SOLUTION: Please help me with this question. A tangent to the parabola x^2 = 16y is perpendicular to the line x - 2y - 3 = 0 .find the equation of this tangent and the coordinate of the po

Algebra ->  Length-and-distance -> SOLUTION: Please help me with this question. A tangent to the parabola x^2 = 16y is perpendicular to the line x - 2y - 3 = 0 .find the equation of this tangent and the coordinate of the po      Log On


   

Question 935929: Please help me with this question.
A tangent to the parabola x^2 = 16y is perpendicular to the line x - 2y - 3 = 0 .find the equation of this tangent and the coordinate of the point of conduct
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
-2y=-x%2B3
y=x%2F2-3%2F2
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%2Bk. You want THIS line to intersect the parabola AT ONE POINT, ....

x%5E2=16%28-2x%2Bk%29
x%5E2=-32x%2B16k
x%5E2%2B32x-16k=0, you need discriminant of this to be 0. This will ensure the intersection will be ONE POINT.

%28-32%29%5E2-4%2A%28-16k%29=0
1024%2B64k=0
64k=-1024
highlight%28k=-16%29

Now you have the line y=-2x-16 and you want to know the intersection of this with the parabola x%5E2=16y.
-
x%5E2=16%28-2x-16%29
x%5E2=-32x-256
x%5E2%2B32x%2B256=0
%28x%2B16%29%5E2=0
THIS means, highlight%28x=-16%29 and the corresponding coordinate y is y=-2%2A%28-16%29-16
y=32-16
highlight%28y=16%29

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.