MATH SOLVE

3 months ago

Q:
# (03.09 MC) Derive the equation of the parabola with a focus at (−2, 4) and a directrix of y = 6. Put the equation in standard form

Accepted Solution

A:

check the picture below.

so the focus point is at -2, 4 and the directrix is at y = 6, now, keeping in mind that the vertex is half-way between those two fellows, from 4 to 6, it'd be the y-coordinate of 5, and therefore, the vertex is at -2,5, as you see there in the picture, and the parabola looks like so. Since the parabola is a vertical one, the squared variable is the "x".

notice the distance "p", is just 1 unit, however, since the parabola is opening downwards, "p" is negative, and thus -1.

[tex]\bf \textit{parabola vertex form with focus point distance}\\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ \boxed{4p(y- k)=(x- h)^2} \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\\\\ -------------------------------\\\\ \begin{cases} h=-2\\ k=5\\ p=-1 \end{cases}\implies 4(-1)(y-5)=[x-(-2)]^2 \\\\\\ -4(y-5)=(x+2)^2\implies y-5=-\cfrac{1}{4}(x+2)^2 \\\\\\ y=-\cfrac{1}{4}(x+2)^2+5[/tex]

so the focus point is at -2, 4 and the directrix is at y = 6, now, keeping in mind that the vertex is half-way between those two fellows, from 4 to 6, it'd be the y-coordinate of 5, and therefore, the vertex is at -2,5, as you see there in the picture, and the parabola looks like so. Since the parabola is a vertical one, the squared variable is the "x".

notice the distance "p", is just 1 unit, however, since the parabola is opening downwards, "p" is negative, and thus -1.

[tex]\bf \textit{parabola vertex form with focus point distance}\\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ \boxed{4p(y- k)=(x- h)^2} \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\\\\ -------------------------------\\\\ \begin{cases} h=-2\\ k=5\\ p=-1 \end{cases}\implies 4(-1)(y-5)=[x-(-2)]^2 \\\\\\ -4(y-5)=(x+2)^2\implies y-5=-\cfrac{1}{4}(x+2)^2 \\\\\\ y=-\cfrac{1}{4}(x+2)^2+5[/tex]