1. Parabola Parameters

Hi

Find the coordinates of 3 points on the parabola x^2 = 4y such that the normals through these 3 points pass through the point (-12,15).

I have no clue whatsoever on how to do this - the previous questions were so different. Please, any help?

2. Originally Posted by xwrathbringerx
Hi

Find the coordinates of 3 points on the parabola x^2 = 4y such that the normals through these 3 points pass through the point (-12,15).

I have no clue whatsoever on how to do this - the previous questions were so different. Please, any help?
any point on the parabola ... $\left(x, \frac{x^2}{4}\right)$

normal slope ... $m = -\frac{2}{x}$

using the point-slope form of a linear equation that passes through $(-12,15)$ ...

$15 - \frac{x^2}{4} = -\frac{2}{x}(-12 - x)$

$15 - \frac{x^2}{4} = \frac{24}{x} + 2$

$0 = \frac{x^2}{4} - 13 + \frac{24}{x}$

$0 = x^3 - 52x + 96$

roots are ...

$x = 2$ , $x = -8$ , $x = 6$

3. Originally Posted by xwrathbringerx
Hi

Find the coordinates of 3 points on the parabola x^2 = 4y such that the normals through these 3 points pass through the point (-12,15).

I have no clue whatsoever on how to do this - the previous questions were so different. Please, any help?
A "normal" to the parabola is perpendicular to the tangent line at that point. The slope of the tangent line at $\left(x_0, \frac{1}{4}x_0^2)\right)$ is $\frac{1}{2}x_0$ so the slope of the normal line is $-\frac{2}{x_0}$. A normal line to the parabola at $\left(x_0, \frac{1}{4}x_0^2\right)$ is $y= -\frac{2}{x_0}\left(x- x_0)+ \frac{1}{4}x_0^3$. Put x= -12, y= 15 into that and solve for $x_0$.