Chapter 10, Problem 4CT

To determine

The equation of the parabola with focus $(-1,4.5)$, vertex $(-1,3)$, and graph the equation

Answer to Problem 4CT

Solution:

The equation of the parabola with focus $(-1,4.5)$, vertex $(-1,3)$ is ${(x+1)}^2=6(y-3)$ and the graph of the equation is

  

Explanation of Solution

Given information:

The parabola has focus $(-1,4.5)$ and vertex $(-1,3)$

The vertex $(-1,3)$ and the focus $(-1,4.5)$ both lie on the vertical line $x=-1$ .

Thus, $x=-1$ is the axis of symmetry

The general form of the parabola is ${(x-h)}^2=4a(y-k)$, opens upwards, if $a>0$ .

Here, $h=-1,k=3$ and $a=1.5$

The distance $a$ from the vertex to the focus is $a=1.5>0$, the parabola opens upwards.

Therefore, the equation of the parabola is

  ${(x-(-1))}^2=4(1.5)(y-3)$

  $\Rightarrow {(x+1)}^2=6(y-3)$ .

To find latus rectum, substitute $y=4.5$ in the equation ${(x+1)}^2=6(y-3)$

  $\Rightarrow {(x+1)}^2=6(4.5-3)$

  $\Rightarrow x^2+2x+1=9$

  $\Rightarrow x^2+2x-8=0$

  $\Rightarrow (x-2)(x+4)=0$

  $\Rightarrow x=2,-4$

The points $(2,4.5)$ and $(-4,4.5)$ determine the latus rectum.

The equation of directrix is $y=k-a$

  $\Rightarrow y=1.5$

By using all the above information, the graph of the parabola ${(x+1)}^2=6(y-3)$ is as follows: