Question

Given the parabola below, find the endpoints of the latus rectum. (x-2)^2=-20(y+2)

Answer 1

The endpoints of the latus rectum are
`(12, -7)` and
`(-8, -7)`.

Explanation:

A parabola with vertex at point
`C(x, y) = (h,k)` and whose axis of symmetry is parallel to the y-axis is defined by the following formula:

`(x-h)^(2) = 4\cdot p \cdot (y-k)` (1)

Where:

`y` - Independent variable.

`x` - Dependent variable.

`p` - Distance from vertex to the focus.

`h`,
`k` - Coordinates of the vertex.

The coordinates of the focus are represented by:

`F(x,y) = (h, k+p)` (2)

The latus rectum is a line segment parallel to the x-axis which contains the focus. If we know that
`h = 2`,
`k = -2` and
`p = -5`, then the latus rectum is between the following endpoints:

By (2):

`F(x,y) = (2, -2-5)`

`F(x,y) = (2,-7)`

By (1):

`(x-2)^(2) = -20\cdot (-7+2)`

`(x-2)^(2) = 100`

`x - 2 = \pm 10`

There are two solutions:

`x_(1) = 2 + 10`

`x_(1) = 12`

`x_(2) = 2-10`

`x_(2) = -8`

Hence, the endpoints of the latus rectum are
`(12, -7)` and
`(-8, -7)`.