# Given the equation (x-3)^2 = 8(y+1) determine: a)  The Vertex V b)  If the parabola opens up, down, left, or right c)  The Focus F

llltkl | Student

The standard form of the equation of a parabola with vertex at (h, k) is given by:

`(x-h)^2=4p(y-k)` (where the axis is vertical) ---- (i)

and `(y-k)^2=4p(x-h)` (where the axis is horizontal) --- (ii)

The given equation is: `(x-3)^2 = 8(y+1)`

This can be written as:

`(x-3)^2 = 4*2*(y+1)`

a) Comparison with the standard form of equation (i) reveals that it is a parabola with vertical axis and having its vertex V at `(3, -1)` .

b) The letter that is squared tells you if the parabola opens up/down or left/right. In fact, it opens along the nonsquared axis. `x^2=4py` parabolas open up and down. Up, if p is positive.

Here p=2, so it opens upwards.

c) The coordinates of the focus F of a parabola, having vertical axis is given by (h, k+p).

The focus F for the given parabola is thus `(3,-1+2)` i.e. `(3,1)`

aruv | Student

Equation of parabola

`(x-3)^2=8(y+1)`

Let us transform the equation in standard form

`X=x-3 and Y=y+1`

So equation reduces to

`X^2=8Y`

Thus

1.vertex of the transform parabola `(X=0,Y=0)`

thus vertex of original parabola is  (x-3=0,y+1=0) i.e.(3,-1)

2. Parabola is defined only if `Y>=0` i.e. `y+1>=0 =>y>=-1`

It means graph opens up.

3. Focus of parabola `(X=0,Y=2)`

`` Thus focus of original parabola is `(x-3=0,y+1=2) i.e. (x=3,y=1)`

It focus at point at (3,1).