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habbster
PARABOLA how to find the vertex and points y^2=9x
done with that x=(1/9)y^2
Okay, now you have your quadratic equation in standard form \[ay^{2}+by+c\] Now the vertex (h,k) is given by \[\huge h=-\frac{b}{2a}\]\[\huge k=\frac{4ac-b^{2}}{4a}\] Alternatively, you can just find h, and then evaluate the function at y = h.
example please I still dont get it
Well, I'm used to them being functions of x, so let's use y = f(x) for now, ok? It's practically the same for y, anyway... so suppose y = x² + 4x + 8 then for this case your a = 1 b = 4 c = 8 Plugging them in your formula gives you h = -2 k = 4 So the vertex of this parabola is at (-2 , 4)
\[h=-0/\frac{ 2}{ 9 }\] i come up with this
Yes, indeed, and finally, you can see that h = ?
Correct. Now you can either compute for k, or just evaluate x when y = 0.
@habbster SOMETHING REALLY IMPORTANT When you have x², then after computing for h and k, your vertex is at (h,k) BUT when you have a y², then after computing for h and k, your vertex is at (k,h) DON'T FORGET!
oh so if the y is the one with a power the vertex is (k,h)
I had to give you the more difficult, yet general formula to prepare you for more complicated parabolas in the future. However, now, you must remember that whenever your quadratic equation is of the form y = kx² or x = ky² in other words, b = 0 and c = 0 Then your vertex is always, ALWAYS at (0,0)
And yes, you're right.
how about the points? where the parabola will pass thru?
Well, just give a random value for y, say m, and solve for x(m). (which is x evaluated when y = m) in that case, the point (x(m), m) is a point on the parabola. You can choose any real number value for y.
For example, I like the number 6. Then x = (1/9)y² x = (1/9)(6)² x = (1/9)(36) x = 4 Hence, the point (4,6) is on the parabola.
if i replace the value of y is equal to 1 then the value of x is 1/9?
That is correct, and hence, the point (1/9 , 1) is on the parabola as well.
ok i get it thank you!! for your time :)
No problem :) Terence out ...
@terenzreignz what is the focus and directx?
I'm really sorry I left too soon... Another form for a parabola is this equation: \[\huge (x - x_{o})^{2} = 4p(y - y_{o})\]or alternatively \[\huge (y - y_{o})^{2} = 4p(x - x_{o})\] First express the equation in this form... (this can usually be done by completing the square, etc...)
That p-value there is key. Anyway, your equation, written like this, takes the form \[\huge (y - 0)^{2} = 4\left( \frac{9}{4} \right)(x - 0)\]or just simply\[\huge y^{2}=4\left( \frac{9}{4} \right)x\] You can see that the value for p is (9/4). Then since this is a horizontal parabola (you have a y²), then your focus and directrix are both (9/4) units away from your vertex along the horizontal. While that may not sound too appealing, it simply means you take the x-value of your vertex, and consider adding and subtracting (9/4) from it. Since your x-value is 0, then adding (9/4) gives you (9/4) and subtracting (9/4) gives you -(9/4). So, copy the y value, you get two points: \[\large \left( \frac{9}{4},0 \right) \ and \ \left( -\frac{9}{4},0 \right)\]. One of these is your focus.
To find out which one, it is necessary to find out in which direction your parabola opens. This particular parabola opens to the right, as the coefficient of y² is positive. This means that your focus is to the RIGHT of your vertex, hence, \[\left( \frac{9}{4},0 \right)\]is your focus.
What about the directrix? It is the line parallel to the y axis passing through the other point we got earlier. So you can see that it is the line \[x = -\frac{9}{4}\]and now you're done.