Question

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

Answer 1

Answer:

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

Step-by-step 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 $(-8, -7)$.