Question

What is the equation of the parabola, in vertex form, with vertex at (2,-4) and directrix y = -6?

Answer 1

Have you considered the definition?

\(\sqrt{(x-2)^2+(y+4)^2} = y + 6\)

Answer 2

no however i still dont understand.

Answer 3

hello @tkhunny

Answer 4

Do the algebra to get that expression onto the standard form.

What I gave is the focus-directrix definition of a parabola.

The left hand side is the distance to the focus.
The right hand side is the distance to the directrix.

For a parabola, these two are equal. Square both sides and simplify. You'll see it.

Answer 5

i have no idea how to solve this still @tkhunny

Answer 6

Did you square both sides? Everything is positive. It will be okay.

Answer 7

so id get (x+2)+(y+4)=(y+6)^2?? @tkhunny

Answer 8

Not quite. Where did the exponents go - the ones that used to be under the radical? You should have:

\( (x-2)^{2}+(y+4)^{2}=(y+6)^2?? \)

And why did you change x-2 to x+2? Don't do that.

Answer 9

oh right sorry.

Answer 10

ok so would the equation be (x-2)^2 =8(y+4) @tkhunny

Answer 11

help @tkhunny

Answer 12

\((x-2)^{2} = (y+6)^{2} - (y+4)^{2}\)

\((x-2)^{2} = y^{2} + 12y + 36 - (y^{2} + 8y + 16)\)

\((x-2)^{2} = 12y + 36 - 8y - 16)\)

\((x-2)^{2} = 4y + 20)\)

\((x-2)^{2} = 4(y + 5))\)

This puts the vertex at (2,-5), which is EXACTLY the middle if the distance between the focus and the directrix.

Hey, look!! We just invented the standard formula!

\((x-h)^{2} = 4p(y + k))\)

With h = 2, k = -5, and p = 1

Answer 13

wow thank you for all your help its been amazing.