Question

Find the equation of the parabaola whose vertex is on the line y-x=2; axis parallel to y-axis, latus rectum is 6 and passing through (-3,-1)Can you please help me? Thanks in advance.:)

Accepted Answer

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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Find the equation of the parabaola whose vertex is on the line y-x=2; axis parallel to y-axis, latus rectum is 6 and passing through (-3,-1) Can you please help me? Thanks in advance.:)

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