Question

what is an equation of a parabola with the given vertex and focus vertex (-2 5) focus (-2 6)

Answer 1

wow another conic problem
4th one tonight, and no one likes them
do you know what this looks like? we need that first

Answer 2

no idea

Answer 3

lets plot the points at least

Answer 4

the fact that the focus is above the vertex should indicate that the parabola opens up

Answer 5

crappy picture but you get the idea i hope

Answer 6

yeah i get it

Answer 7

general form will be something like
\[4p(y-k)=(x-h)^2\] we know the \(x\) is squared because it opens up
now all you need is \(p\) and \((h,k)\)

Answer 8

the vertex is \((-2,5)\) making \(h=-2,k=5\)

Answer 9

so now \[4p(y-k)=(x-h)^2\]looks like
\[4p(y-5)=(x+2)^2\]

Answer 10

all you need is \(p\) which is the distance from the vertex to the focus
the vertex is \((-2,5)\) and the focus is \((-2,6)\) which is one unit up, making \(p=1\) and the whole thing is therefore
\[4(y-5)=(x+2)^2\]

Answer 11

wow that was honestly really simple... i feel silly now lol. thank you

Answer 12

simple once you do two or three, yes
yw