Question

Can someone please help me find the focus and directrix of y^2+6y-2x+13=0?

Answer 1

I've been told that the directrix is h-a/4... if that is correct I got x=-1. I dont know how to find the focus of this equation though.

Answer 2

@ranga can you please help me with this?

Answer 3

This is a sideways parabola because y is squared and x is not.

First put it in the vertex form: x = a(y - k)^2 + h
(by completing squares)

Answer 4

what numbers do i plug into what numbers?

Answer 5

y^2+6y-2x+13 = 0
isolate the x by adding 2x to both sides:
y^2 + 6y +13 = 2x
divide both sides by 2
1/2 * (y^2 + 6y +13) = x or
x = 1/2 * (y^2 + 6y +13)
complete the square for (y^2 + 6y +13)

Answer 6

I got y=-3+-2i is that correct?

Answer 7

To complete the square, take the coefficient of y, divide it by 2.
This is the number that will go inside the square and you will have to subtract the square of that number.
Coefficient of y is 6. 6/2 = 3. So 3 will go inside the square and 3^2 should be subtracted:
y^2 + 6y +13 = (y + 3)^2 - 9 + 13 = (y + 3)^2 + 4
So x = 1/2 * { (y + 3)^2 + 4 }
distribute the 1/2
x = 1/2(y + 3)^2 + 2

Answer 8

oh so y=-3+-2i is wrong?

Answer 9

yes. we are not solving quadratic equations here. We are completing the square to put the parabola equation in vertex form.

Answer 10

would it be x=1/2(y^2+6y+13) ?

Answer 11

That is what we started with: x=1/2(y^2+6y+13)
But the expression within the parenthesis:
We have to complete the square which is what I did above.

Answer 12

so x = 1/2(y + 3)^2 + 2 is the completed square?

Answer 13

We have to make x=1/2(y^2+6y+13) look like this: x = a(y - k)^2 + h

Answer 14

yes. x = 1/2(y + 3)^2 + 2 looks like x = a(y - k)^2 + h

Answer 15

oh okay, so what is my next step?

Answer 16

we can compare and determine a = 1/2; k = -3; h = 2
The vertex is (h,k). So the vertex for this parabola is (2, -3)

Answer 17

so the length of the foacl cord would be 1/2 or would it be 2?

Answer 18

The focus is at distance "p" from the vertex where p = 1/(4a)

We got a = 1/2.
So p = 1/(4*1/2) = 1/2

The focus is at distance 1/2 from the vertex.

This is a parabola that is sideways and so the distance of the focus from the vertex has to be added to the x-coordinate. Vertex at (2,-3) So the focus is at: (2+1/2, -3)
(2.5, -3)

Answer 19

how would i find the directrix? I'm also still a tad bit confused on what the length of the focal chord is. would the focal chord be 1/2 or 2?

Answer 20

half.

Answer 21

The directrix will be on the other side of the vertex at the same distance as focus is from the vertex. * * *
directrix vertex focus

Answer 22

subtract 1/2 from the x-coordinate of the vertex. vertex at (2, -3)
directrix at x = 2-1/2 = 1.5
x = 1.5 is the equation of the directrix.

Answer 23

so both the directrix and the length of the focal chord is 1/2? because i thought that the length of the focal chord was equal to a which in this case would be 2, right?

Answer 24

no. if p is the distance between vertex and focus (or vertex and directrix) then p = 1/4a.

where a is the coefficient of the y^2 term.