Question

How do you find an equation of the parabola with focus $\left( 2,2 \right)$ and directrix $x=-2$?

Answer 1

Hint: We equate the given information about the parabolic curve with focus $\left( 2,2 \right)$ and directrix $x=-2$. We take the general equation of \[{{\left( y-\beta \right)}^{2}}=4a\left( x-\alpha \right)\] . We find the number of x intercepts and the value of the y intercept. We also find the value of $a$ from the coordinates of the focus to place the curve in the graph.

Complete step-by-step answer:
The general equation \[{{\left( y-\beta \right)}^{2}}=4a\left( x-\alpha \right)\] is a parabolic curve.
For the general equation $\left( \alpha ,\beta \right)$ is the vertex. 4a is the length of the latus rectum. The coordinate of the focus is $\left( \alpha +a,\beta \right)$. The equation of the directrix for $x+a=\alpha $. The distance between the focus and the directrix is 2a.
We now equate it with the given information of the parabola with focus $\left( 2,2 \right)$ and directrix $x=-2$. So, we get $\beta =2,\alpha +a=2$.
We also have that for the equation of the directrix $x=\alpha -a=-2$.
We add these equations to get $\alpha +a+\alpha -a=2-2$ which gives $\alpha =0$.
Therefore, the value of a is $a=\alpha +2=2$.
We put these values in the equation to get
 \[\begin{align}
  & {{\left( y-2 \right)}^{2}}=4\times 2\left( x-0 \right) \\
 & \Rightarrow {{\left( y-2 \right)}^{2}}=8x \\
\end{align}\]
The simplified form is \[{{y}^{2}}-4y-8x+4=0\] .
Therefore, the equation of the parabola with focus $\left( 2,2 \right)$ and directrix $x=-2$ is \[{{y}^{2}}-4y-8x+4=0\] .

Note: The leftmost point of the function \[{{y}^{2}}-4y-8x+4=0\] is $\left( 0,2 \right)$, the vertex of the curve. The graph is bounded at that point. But on the other side the curve is open and not bounded. The general case of parabolic curve is to be bounded at one side to mark the vertex.