Question

Derive the equation of the parabola with a focus at (-7, 5) and a directrix of y = -11.

Answer 1

The parabola derived from a focus and directrix is the set of points that satisfies the condition that their distance from the focus is equivalent to the vertical distance from the directrix. So, we have \[\sqrt{(-7-x)^2+(5-y)^2}=y+11\] Squaring both sides gives us \[(x^2+14x+49)+(25-10y+y^2)=y^2+22y+121\] SImplifying, we get\[x^2+14x+49=32y+121-25\] Simplifying further gives us \[x^2+14x-47=32y\] so our final equation is \[y=\frac{x^2+14x-47}{32}\]

Answer 2

Although that doesn't match any of the listed answers, I'm fairly sure it's correct or I'm misreading the problem; Wolfram|Alpha agrees.

Answer 3

@HarryPotter5777 it's correct, though that's not how they are typically taught

Answer 4

The first thing is to make a quick sketch. Did you do that?

Answer 5

Generic equation for a parabola:
(x−h)^2=4p(y−k)

the coordinates of the vertex are:
Vertex−(h,k)

The Coordinates of the vertex are:
Focus−(h,(k+p))

The equation for the directix:
y=k−p


Therefore your first question - Focus = (-7,5), Directix = -11
From the equation for a directix:
−11=y=k−p

From the y coordinate for the focus:
5=k−p

You can then solve these equations simultaneously and get k and p:
k = -3, p = 8 you know from the x coodinate of the focus is h = -7 so you can wright the equation for the parabola
(x−(−7))^2=4×8(y−(−3))
Simplifies to :
(x+7)^2=32(y+3)
y=1/32(x+7)^2-3

Answer 6

i think A is correct one.

Answer 7

this is my second one

Answer 8

You're not gonna be able do the rest of the homework if you're simply just looking for answers

Answer 9

A is correct. Taking @harrypotter5777's answer:

\[y = \frac{1}{32}(x^2+14x−47)\]We can complete the square on the quadratic by adding 47+49=96 inside the parentheses and subtracting 96/32=3 outside:
\[y = \frac{1}{32}(x^2+14x-47 + 96 - 96) = \frac{1}{32}(x^2+14+49) -\frac{96}{32}\]\[y=\frac{1}{32}(x+7)^2-3\]which matches A. A graph showing the various details is attached.

Answer 10

Your other problem

If the distance between directrix and focus is d, then the point d units to the right of the focus will be on the parabola. This allows us to use vertex form:
\[y = a(x-h)^2+k\]We know the vertex is at \((h,k) = (0,(-1+1)/2) = (0,0)\) because the vertex is midway between focus and directrix. Therefore \((0,0)\) is a point on the parabola, and we can simplify our equation to \[y = ax^2\]Now to find \(a\), we use our other point whose location we have determined, and that is \((0+2,-1+2) = (2,1)\). That is the point which is \(d =2\) units to the right of the focus.

\[y = ax^2\]\[1 = a(2)^2\]Solve for \(a\) and plug into your equation for the final formula.