Can you please help me with the following questions.

Graph the parabola with the given equation.

(y-2)^2=6(x-1)

I know the vertex would be (1,2)

(x+1)^2=6(y-1)

The vetex would be (-1,1)

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- Nov 23rd 2007, 07:31 PMsoly_solParabolas- Need help
Can you please help me with the following questions.

Graph the parabola with the given equation.

(y-2)^2=6(x-1)

I know the vertex would be (1,2)

(x+1)^2=6(y-1)

The vetex would be (-1,1) - Nov 23rd 2007, 08:24 PMkalagota
yes, and where should be the focus?

note that the distance of the focus from the vertex of the parabola $\displaystyle (y-k)^2 = 4p(x-h)$ is p. what is your p in your equation?

also note that the distance of the endpoints of the latus rectum is 4p, or teh distance of one endpoint from the focus is 2p.. (by the way, remember that the latus rectum is the segment perpendicular to the axis of symmetry of the parabola passing through the focus)

with this information, i hope you could sketch the graphs..

- Nov 24th 2007, 11:05 AMsoly_sol
Graph the parabola with the given equation.

(y-2)^2=6(x-1)

I know the vertex would be (1,2)

Here else is what I got:

I got p=24

foci: (25,2)

directrix: x=-23

axis:y=2

(x+1)^2=6(y-1)

The vetex would be (-1,1)

foci: (-1, 25)

directrix: y=-23

axis: -1 - Nov 24th 2007, 03:22 PMkalagota