# 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.:)

### 1 Answer | Add Yours

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).` **