Find the focus of the parabola that has vertex (−3, 2), opens horizontally, and passes through the point (−10, 1).
(−83/28, 2)(−87/28, 2) (−85/28, 2) (−4, 2)None of the above
Find the focus of the parabola that has vertex (−3, 2), opens horizontally, and passes through the point (−10, 1).
(−83/28, 2)(−87/28, 2) (−85/28, 2) (−4, 2)None of the above
I almost didn't view this thread, because your various replies made it look as though it had been answered already!
To learn the basic process for finding the equation from some provided information, try here.
You know that the equation is something along the lines of 4p(x - h) = (y - k)^2, because the parabola opens horizontally instead of vertically.
You are given that (h, k) = (-3, 2). You are also given that the point (-10, 1) lies on the curve. Then:
. . . . .4p(x + 3) = (y - 2)^2
Plugging in the given point:
. . . . .4p(-10 + 3) = (1 - 2)^2
. . . . .4p(-7) = (-1)^2
. . . . .-28p = 1
. . . . .p = -1/28
Since p is negative, then the parabola opens to the left, with the vertex being on the right. To which side of the vertex, then, is the focus? How far is the focus from the vertex?
Then what are the coordinates of the focus?
If you get stuck, please reply showing all of your work and reasoning so far.
Thank you!
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