Chapter 11, Problem 37RE

To determine

To find: The type of curve and to find the focus, vertices and to plot the equation graph.

Answer to Problem 37RE

Through the Standard equation of parabola when vertices are $(h,k)$ , the type of curve given is found to be a parabola with:

Co-ordinates of vertices $\left(-64,8\right)$ .

Co-ordinates of focus $\left(-\frac{255}{4},8\right)$

Explanation of Solution

Given information:

The equation of the curve is:

  $x=y^2-16y$

Calculation:

This is the equation of a parabola whose center is not at origin $(0,0)$ . Standard equation of parabola when vertices are $(h,k)$ is given by:

  ${(y-k)}^2=4a(x-h)\to 1$

Here axis of symmetry is $y=k$ and focus = $\left(h+a,k\right)$

  $x=y^2-16y$

Rearrange the above equation,

  $\begin{array}{l}{y}^2-16y=x\\ y^2-16y+64-64=x\\ {(}^y=(x+64)\to 2\end{array}$

Comparing equation $1$ and $2$

Vertices are $=\left(-64,8\right)$

  $a=\frac{1}{4}$

Focus $=\left(-64+\frac{1}{4},8\right)$

  $=\left(-\frac{255}{4},8\right)$

Graph:

  

Interpretation :

Through the Standard equation of parabola when vertices are $(h,k)$ , the type of curve given is found to be a parabola with:

Co-ordinates of vertices $\left(-64,8\right)$ .

Co-ordinates of focus $\left(-\frac{255}{4},8\right)$