Question

I did the work. I just need help putting it in standard form.

Derive the equation of the parabola with a focus at (4, −7) and a directrix of y = −15. Put the equation in standard form.

My answer in vertex form:
1/16(x-4)^2-7

Answer 1

your equation has the vertex at 4, -7
But you want a parabola that has its focus at 4,-7

I think the vertex is half way between the focus and the directrix.

Answer 2

My vertex was (4,-11).

Answer 3

\(\bf \cfrac{1}{16}(x-\color{red}{4})^2\color{red}{-7}\) vertex

Answer 4

Focus: (4,-7)
Directrix: y= -15
Vertex: (4,-11)
Distance: 4

Answer 5

well, that's now what yours shows

Answer 6

yes. so you want -11 in your equation not -7

Answer 7

that's not rather

Answer 8

Here are my choices:

f(x) = x2 − 8x + 11

f(x) = x2 − 8x − 10

f(x) = x2 −1/2 x + 11

f(x) = x2 −1/2 x − 10

Right now, I think the answer is the first one.

Answer 9

you should start with

My answer in vertex form:
1/16(x-4)^2- 11

Answer 10

Okay, how do I get from that to standard form?

Answer 11

multiply out (x-4)(x-4)

Answer 12

1/16[(x-4)(x-4)]-11

Answer 13

multiply (x-4)(x-4) to get x^2 -4x -4x + 16 = x^2 -8x +16
so you have
\[ \frac{ x^2}{16} - \frac{8}{16}x + \frac{16}{16} - 11 \]
that simplifies to
\[ \frac{ x^2}{16} - \frac{1}{2}x + 10\]

Answer 14

*\[\frac{ x^2}{16} - \frac{1}{2}x - 10 \]

Answer 15

all of your choices left out the divide by 16 on x^2

Answer 16

Okay, Thanks so much!