How do you find an equation of a parabola given endpoints of latus rectum are (-2,-7) and (4,-7)?

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How do you find an equation of a parabola given endpoints of latus rectum are (-2,-7) and (4,-7)?

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Answer 1 · 2017-08-03T20:38:50

The vertex form of a parabola of this type is:

$y = \frac{1}{4 f} {(x - h)}^{2} + k [1]$ $y = 1/(4f)(x-h)^2+k" [1]"$

where $(h, k)$ $(h,k)$ is the vertex and $f = y_{focus} - k$ $f = y_"focus"-k$ .

The x coordinate of the vertex, h, is the midpoint between the x coordinates of the two points:

$h = \frac{4 + (- 2)}{2} = 1$ $h = (4+ (-2))/2 = 1$

$y = \frac{1}{4 f} {(x - 1)}^{2} + k [2]$ $y = 1/(4f)(x-1)^2+k" [2]"$

We know that $4 f$ $4f$ is $\pm$ $+-$ the length of the latus rectum:

$4 f = 4 - (- 2)$ $4f = 4 - (-2)$ or $4 f = - 2 - 4$ $4f = -2 -4$

$4 f = 6$ $4f = 6$ or $4 f = - 6$ $4f = -6$

We are not told whether the parabola opens up or down and you have only asked for one of the two possible equations, therefore, I shall choose the positive value:

$y = \frac{1}{6} {(x - 1)}^{2} + k [3]$ $y = 1/6(x-1)^2+k" [3]"$

We can find the value of k substituting in one of the points:

$- 7 = \frac{1}{6} {(4 - 1)}^{2} + k$ $-7 = 1/6(4-1)^2+k$

$- 7 = \frac{1}{2} + k$ $-7 = 1/2+k$

$k = - 7.5$ $k = -7.5$

$y = \frac{1}{6} {(x - 1)}^{2} - 7.5 \leftarrow$ $y = 1/6(x-1)^2-7.5 larr$ answer.

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How do you find an equation of a parabola given endpoints of latus rectum are (-2,-7) and (4,-7)?

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