Chapter 2.3, Problem 52E

To determine

To find: An equation of each described parabola and to identify the directrix of each parabola.

Answer to Problem 52E

The equations of the parabolas must be:

  $\begin{array}{}$ y=\frac{1}{8}{(x-a)}^2+b-2$$ x=\frac{1}{8}{(y-b)}^2+a-2$\end{array}$

The directrices of the parabolas are:

  $\begin{array}{l}y=b-2-2=b-4\\ \&x=a-2-2=a-4\end{array}$ respectively.

Explanation of Solution

Given information: The two parabolas have same focus $(a,b)$ and focal length of $2units$ .

Calculation:

Standard equations of a parabola with vertex at $(h,k)$ are:

  $y=\frac{1}{4p}{(x-h)}^2+k$ and $x=\frac{1}{4p}{(y-k)}^2+h$ with foci at $(h,k+p)$ , $(h+p,k)$ and directrices at $y=k-p,x=h-p$ respectively.

Since the two parabolas have same focus $(a,b)$ and focal length of $2units$ ,

So comparing with general form, the first parabola has the vertex $(a,b-2)$ and second has the vertex $(a-2,b)$ .

Thus, the equations of the parabolas must be:

  $\begin{array}{}$ y=\frac{1}{8}{(x-a)}^2+b-2$$ x=\frac{1}{8}{(y-b)}^2+a-2$\end{array}$

The directrices of the parabolas are:

  $\begin{array}{l}y=b-2-2=b-4\\ \&x=a-2-2=a-4\end{array}$ respectively.