__Category__: Analytic Geometry, Algebra"Published in Newark, California, USA"

Find the vertex, focus, and end points of the latus rectum for parabola with an equation

and sketch the graph.

__Solution__:

The first thing that we have to do is to write the equation of a parabola in standard form as follows

Transpose -3y and 10 to the right side of the equation

Complete the square at the left side of the equation by adding 1 on both sides of the equation

Therefore, the vertex of a parabola is

**V(1, 3)**.

Since the second degree in the equation is x, the parabola is either concave upward or downward.

Next, let's solve for the coordinates of the focus and the ends points of the latus rectum. Since 4a = 3 is positive, then the parabola is concave upward. a is the distance between the focus and the vertex of a parabola. Solve for the value of a as follows

4a = 3

a = ¾

To get the coordinates of the focus, count ¾ units upward from the vertex. Therefore, the coordinates of the focus is

**F(1, 3¾)**.

To get the coordinates of the first end point of the latus rectum, count 2a = 2(¾) = 1½ units to the right from focus. The first end point of the latus rectum is

**B(2½, 3**

**¾)**.

To get the coordinates of the other end point of the latus rectum, count 2a = 2(¾) = 1½ units to the left from focus. The other end point of the latus rectum is

**A(-**

**½, 3**

**¾)**.

Finally, we have the coordinates of the vertex, focus, and the end points of the latus rectum. We can sketch the graph of the parabola as follows

Photo by Math Principles in Everyday Life |