Question

In two or more complete sentences, Explain how you would find the equation of a parabola, given the coordinate of the focus and the equation of the directrix.  

Graph and describe the elements of

Begin by finding the value of p:

If p = 4, then the equation of the directrix is y = -(4) → y = -4. The coordinate of the focus is (0, 4) and the vertex is (0, 0). The axis of symmetry is the y-axis, x = 0.

Example 3:

Graph and describe the elements of - 24y = x2.

Begin by putting the equation into standard form and solve for y:

Now, solve for p:

If p = -6, then the equation of the directrix is y = -(-6) → y = 6. The coordinate of the focus is (0, -6) and the vertex is (0, 0). The axis of symmetry is the y-axis, x = 0.

Answer 1

so .. .if you check the picture below.. .those are both of them

so.. let's see the one on the left-side, "p" is 4, the distance from the focus point to the vertex, now, in this case, the vertex is given, if it wasn't, say we know the focus point is at 0,4 and the directrix is y = -4, well, the focus point is "inside" the parabola, so, that means the parabola opens upwards and is vertical

where's the vertex? well, "p" is the distance from the vertex to either, so from 0,4 down to -4 over the y-axis, is 8units, half-way in between, is the vertex, axis of symmetry is x = 0, so, moving down the axis of symmetry, from 0,4 down 8/2 or 4 units, we find the vertex, 4-4=0, so is at 0,0

or you can also get it from the directrix, from -4 up 4 units -4 + 4 = 0
so, 0 for the "k" value, and the "h" value is just the axis of symmetry or x = 0, h = 0 then

now the equation is just a matter of plugging in the values


`\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}\\\\ -----------------------------\\\\ \begin{cases} p=4\\ h=0\\ k=0 \end{cases}\implies (x-0)^2=4(4)(y-0)\implies x^2=16y \\\\\\ \boxed{\cfrac{1}{16}x^2=y}`

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now, for the second one, the one on the right-side in the picture

since the directrix is above the focus point, that means, is a vertical parabola and it opens downwards, if it opens downwards, it means "p" is negative, we know "p" is 6 units, but since the parabola is going down, is -6 then


`\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}\\\\ -----------------------------\\\\ \begin{cases} p=-6\\ h=0\\ k=0 \end{cases}\implies (x-0)^2=4(-6)(y-0)\implies x^2=-24y \\\\\\ \boxed{-\cfrac{1}{24}x^2=y}`