You need to remember the equation of parabola such that:
`(y - k)^2 = 4p(x - h)`
`(h,k)` represents the vertex of parabola
`4p` represents latus rectum
Since the problem provides the information that latus rectum is 6, hence you need to solve the following equation such that:
`4p = 6 => p = 6/4 => p = 3/2`
Since the problem provides the information that parabola passes through (-3,-1), you need to substitute -3 for x and -1 for y such that:
`(-1 - k)^2 = 6(-3 - h) `
You need to expand the square such that:
`1 + 2k + k^2 = -18 - 6h`
You also know that the vertex of parabola is on the line `y-x=2` , hence, you need to substitute h for x and k for y in equation of the support line such that:
`k - h = 2 => k = 2 + h`
You need to substitute `2+h` for k in equation `1 + 2k + k^2 = -18 - 6h` such that:
`1 + 2(2+h) + (2+h)^2 + 18 + 6h = 0`
`1 + 4 + 2h + 4 + 4h + h^2 + 18 + 6h = 0`
`h^2 + 12h + 27 = 0`
You need to find `h_(1,2)` using quadratic formula such that:
`h_(1,2) = (-12 +- sqrt(144 - 108))/2`
`h_(1,2) = (-12 +- sqrt36)/2`
`h_(1,2) = (-12 +- 6)/2 => h_1 = -3 ; h_2 = -9`
You may find `k_(1,2)` such that:
`k_1 = 2-3 => k_1 = -1`
`k_2 = 2-9 => k_2 = -7`
Hence, evaluating the equations of parabolas under the given conditions yields `(y+1)^2 = 6(x +3)` and `(y+7)^2 = 6(x +9).`
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