Question

Derive the equation of the parabola with a focus at (−5, 5) and a directrix of y = −1.

Answer 1

Answer: The equation of parabola is $(x+5)^2=12(y-2)$.

Explanation:

It is given that the focus of the parabola is (−5, 5) and a directrix of y = −1.

The standard form of a parabola is,

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

Where, (h,k+p) is the focus and y=k-p is the directrix.

It is given that  the focus of the parabola is (−5, 5).

$(h,k+p)=(-5,5)$

On comparing both sides we get,

$h=-5$

$k+p=5$      .... (1)

The directrix of y = −1.

$k-p=-1$       .... (2)

Add equation (1) and (2),

$2k=4$

$k=2$

Put this in equation (1),

$p=3$

Put p=3, k=2 and h=-5 in standard equation of parabola.

$(x-(-5))^2=4(3)(y-2)$

$(x+5)^2=12(y-2)$

Therefore, the equation of parabola is $(x+5)^2=12(y-2)$.

Answer 2

Check the picture below.

so, the left-part of the picture is the focus point and the directrix... .so.. notice, the focus is above the directrix, meaning is a vertical parabola and is opening upwards.

now, looking at the right-part of the picture, bear in mind that, the vertex is "p" units away from the directrix and the focus, so, the vertex is half-way between those fellows, notice in the picture what "p" is, keep in mind that, because the parabola is opening upwards, "p" is a positive unit, thus is 3.

$\bf \textit{parabola vertex form with focus point distance}\\\\ \begin{array}{llll} (y-{{ k}})^2=4{{ p}}(x-{{ h}}) \\\\ \boxed{(x-{{ h}})^2=4{{ p}}(y-{{ k}}) }\\ \end{array} \qquad \begin{array}{llll} vertex\ ({{ h}},{{ k}})\\\\ {{ p}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\\\\ -------------------------------$

$\bf \begin{cases} h=-5\\ k=2\\ p=3 \end{cases}\implies [x-(-5)]^2=4(3)(y-2) \\\\\\ (x+5)^2=12(y-2)\implies \cfrac{(x+5)^2}{12}=y-2\implies \cfrac{(x+5)^2}{12}+2=y \\\\\\ \cfrac{1}{12}(x+5)^2+2=y$