### PARABOLA

Hey guys

I have nothing to say so im just going stray 2 explaining what happen 2day. But the thing is we did not do much today, it was just 2 question we had to work it out 2gether and play with this circle Mr. K gave it 2 us. Here are the 2 questions:

1) A parabola has vertex (2,-3).

And contains the point (9,-10).

Find its equation and sketch the graph if it is oriented:

a) Vertically

(x-h)^2=4p(y-k)

(x-2)^2=4p(y+3)

(9-2)^2=4p(-10+3)

(7)^2=4p(-7)

49=-28p

-7/4=p

Therefore the equation is (x-2)^2=-7(y+3)

And the graph looks like this:

The point inside the parabola is the focus point which is (2,-19/4)

The line on top of the parabola is axis of symmetry which is Y=-5/4

b) Horizontally

(y-k)^2=4p(x-h)

(-10+3)^2=4p(9-2)

49=28p

49/28=p

7/4=p

Therefore the equation is (y+3)^2=7(x-2)

And the graph looks like the first graph with the same vertex but the graph opens right, the focus is (15/4,-3), and the axis of symmetry is X=1/4.

Anyways i think this was it for today's class, so the next scribe for tomorrow is

**a-b33-l**

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