Grove Prelim Book 2U 11.7 Question 1

$y=x^2/12$

$y'=x/6$

At x=6, $y'=6/6=1$

Normal gradient is $m=-1/1=-1$

$y-3=-1(x-6)$

$y-3=-x+6$

$y=-x+9$

Solve 
$y=-x+9$  (1)
$x^2=12y$   (2)

(1) x 12:
$12y=-12x+108$   (1)

so (1) subbed into (2) gives

$x^2=-12x+108$

$x^2+12x-108=0$                  

                  sandbox:  108=54 x 2 = 27 x 2^2 = 3^3 x 2^2
                                                                           3x2=6, 3^2 x 2=18!! yes

$(x+18)(x-6)=0$

$x=-18$ is where the normal cuts the parabola again.

$y=- -18 +9=27$

$P(-18,27)$