# Thread: Need help with a parametrics question

1. ## Need help with a parametrics question

Hi everyone!

The points A (−6, -9) and B(u,v) lie on the parabola x^2 =-4ay such that AB is a normal to the parabola at B.
(i) Find the value of a.
(ii)Given the roots of x^3 − 28x + 48 = 0 are −6, 2, 4, find the coordinates of B.

I can't seem to get part (ii).

2. ## Re: Interesting parametrics question!

i)

Using the general equation of the parabola

$x^2=-4ay$

and the given point $A(-6,-9)$, we can deduce $a=1$.

ii)

Let $B$ be given by

$B\left(u, -\frac{u^2}{4}\right)$

The cartesian equation of the normal to the parabola is

$y-y_1 = -\frac{2a}{x_1}(x-x_1)$.

Using the coordinates of $B$ to substitute in for $x_1,\,y_1$,

$y+\frac{1}{4}u^2 = -\frac{2}{u}(x-u)$

Which gives the equation of the normal to the parabola at the point $B$.

Because we require $A$ to be on the normal, we can substitute $A(-6,-9)$ into the equation and solve for $u$.

After substitution and simplfication you will get the equation given to you in ii).

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