Question

FIND THE FOCUS OF THE PARABOLA.
-x^2-4x+3y-13=0
************please help*********

Answer 1

Separate out the y and write the function as y = ....
Write the equation of the parabola in vertex form.
Once you do that we can find the focus.

Answer 2

yes i did that... 3y=x^2+4x=13 then i divided everything by 3 and it got confusing @ranga

Answer 3

Put in this form:
\(y = ax^2 + bx + c \), where,
$$
a = \frac{1}{4p}; \ \ b = \frac{-h}{2p}; \ \ c = \frac{h^2}{4p} + k; \ \
$$
The vertex is at the coordinate <h,k>:
$$
(x - h)^2 = 4p(y - k) \
$$
So solve the 1st set of equations for \(p,h\) and \(k\). To find the focus, add \(<0,p>\) to the vertex, \(<h,k>\)

Answer 4

See http://en.wikipedia.org/wiki/Parabola

Answer 5

tamilanda...

Answer 6

yes i get it but i get confused only with this problem b/c of the fraction @ybarrap @mastrblastr

Answer 7

-x^2-4x+3y-13=0
3y = x^2 + 4x + 13
divide by 3
y = 1/3(x^2 + 4x + 13)
We did not divide each term by 3 because when completing square we want the x^2 coefficient to be 1 and so we will leave 1/3 as a factor outside.

Can you complete the square of (x^2 + 4x + 13) ?