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
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)`
Equation of parabola
Let us transform the equation in standard form
`X=x-3 and Y=y+1`
So equation reduces to
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).