Question

A parabola can be drawn given a focus of (10, 7) and a directrix of x = 6 Write the equation of the parabola in any form.

Answer 1

Check the picture below, so the parabola looks more or less like so, with a positive "p" distance of 2, with the vertex half-way between the directrix and the focus point.

`\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`

`\begin{cases} h=8\\ k=7\\ p=2 \end{cases}\implies 4(2)(~~x-8~~) = (~~y-7~~)^2 \implies 8(x-8)=(y-7)^2 \\\\\\ x-8=\cfrac{1}{8}(y-7)^2\implies {\Large \begin{array}{llll} x=\cfrac{1}{8}(y-7)^2+8 \end{array}}`

Answer 2

The standard form of the equation of a parabola with a vertical axis of symmetry is:

(y - k)^2 = 4p(x - h)

where (h, k) is the vertex of the parabola and p is the distance from the vertex to the focus and to the directrix.

The directrix is a horizontal line, so the vertex is (6, 7). The distance from the vertex to the focus is 4 units, since the focus is 4 units above the vertex. Therefore, p = 4.

Substituting the values of h, k, and p into the standard form equation, we get:

(y - 7)^2 = 16(x - 6)

Expanding the right side and rearranging, we get:

y^2 - 14y + 49 = 16x - 96

16x = y^2 - 14y + 145

Dividing both sides by 16, we get:

x = (1/16)y^2 - (7/8)y + 9.0625

Therefore, the equation of the parabola is x = (1/16)y^2 - (7/8)y + 9.0625.