First exercise
Finding the foci and vertice of (Y-2)² = 4(X-3)


Second exercise:
Equation of hyperbola F(± 4, 0) and K= 6

I would deeply appreciate any quick prompt answer. Thanks.

Quote:

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Originally Posted by Danne1

First exercise
Finding the foci and vertice of (Y-2)² = 4(X-3)

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Note that you can't cut and paste math symbols from other applications into the question box because it becomes unreadable. But if you mean this:

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

you have a parabola that is laying on its side. The general form for the parabola is 4p(x-x1) = (y-y1)^2 where (x1,y1) is the vertex and 'p' is the distance between focus and vertex. So here p = 1; can you finish it from here?

Quote:

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Originally Posted by Danne1

Second exercise:
Equation of hyperbola F(± 4, 0) and K= 6.

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I assume that you mean is the foci are at (4,0) and (-4, 0). From this you know that c = 4 (distance from the center to either focus). Not sure what "k" means, but if you mean eccectricity (e) then since c = ae that tells you that the value for 'a' is 4/6 = 2/3. The value for b can be found from c = sqrt(a^2+b^2).