Question

Write the equation of a parabola whose directrix is X=11.25 and focus is (6.75,6) Please Help me

Answer 1

Check the picture below.  So the parabola looks like so.

we know the directrix is at x = 11.25, the blue line, and we also know the focus point is at (6.75 , 6), meaning that the focus point is to the left-side of the directrix, it means is a horizontal parabola, thus the squared variable is the "y", and it also means that our "p" distance will be negative.

we know the the vertex is half-way between the directrix and the focus point, so 11.25 - 6.75 = 4.5, half that is 2.25 units, so if we move from either the focus point or the directrix, 2.25 units towards the other, we'll get the vertex, well, hmmm say 11.25 - 2.25 = 9, and the y-coordinate is of course the same as the focus point's.

now, since "p" is negative, that means p = -2.25.

$\bf \textit{horizontal parabola vertex form with focus point distance} \\\\ 4p(x- h)=(y- k)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h+p,k)}\qquad \stackrel{directrix}{x=h-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\supset}\qquad \stackrel{"p"~is~positive}{op ens~\subset} \end{cases} \\\\[-0.35em] ~\dotfill$

$\bf \begin{cases} h = 9\\ k = 6\\ p = -2.25 \end{cases}\implies 4(-2.25)(x-9)=(y-6)^2\implies -9(x-9)=(y-6)^2 \\\\\\ x-9=-\cfrac{1}{9}(y-6)^2\implies x=-\cfrac{1}{9}(y-6)^2+9$