Question

Find the standard form of the equation of the parabola with a vertex at the origin and a focus at (0, -7).

Answer 1

The standard form of the equation of a parabola with a vertex at the origin and a focus at (0, -7) is y = (1/28)x^2.

Step-by-step explanation:

The question asks for the standard form of the equation of a parabola with a vertex at the origin (0,0) and a focus at (0, -7). In general, the equation of a parabola with a vertical axis of symmetry and vertex at the origin can be written as:

y = ax^2

Since the focus is at (0, -7), the directrix is at y = 7. The distance from the vertex to the focus, which is also the distance from the vertex to the directrix, is p = 7. Therefore, our coefficient a is 1/(4p) or 1/(4*7). This gives us a value of 1/28. Thus, the equation of the parabola is:

y = (1/28)x^2

This is the standard form of the equation of the parabola with the given vertex and focus.

Answer 2

so hmm check the picture below

we know the vertex is at the origin, and the focus point is below it, that means two things, the parabola is vertical and it's opening downwards

notice the distance "p", from the vertex to the focus point, is just 7 units, however, since the parabola is opening downwards, the "p" value will be negative, so p = -7


`\bf \textit{parabola vertex form with focus point distance}\\\\ \begin{array}{llll} (y-{{ k}})^2=4{{ p}}(x-{{ h}}) \\\\ \boxed{4{{ p}}(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}\\\\ -------------------------------\\\\ 4{{ p}}(y-{{ k}})=(x-{{ h}})^2\quad \begin{cases} h=0\\ k=0\\ p=-7 \end{cases}\implies 4(-7)(y-0)=(x-0)^2 \\\\\\ -28y=x^2\implies y=-\cfrac{1}{28}x^2`