Chapter 9, Problem 1CT

To determine

To find: Vertices and Foci of given conic.

Answer to Problem 1CT

Conic is Parabola

Focus of parabola = $\left(2,0\right)$

Vertex of parabola = $\left(0,0\right)$

Explanation of Solution

Given information:

  $y^2-8x=0$

  $y^2-8x=0$

Graph of given Conic is

  

Here red line represents the given conic

So, from the graph we can say thatgiven conic is Parabola If parabola in the form $y^2=4ax$

Then

Foci of parabola = $\left(a,0\right)$

Vertex of parabola= $\left(0,0\right)$

Given

  $y^2-8x=0$

  $\Rightarrow y^2=8x$

Comparing with general equation $y^2=4ax$

We get

  $\begin{array}{l}\Rightarrow 4a=8\\ \Rightarrow a=\frac{8}{4}\\ \Rightarrow a=2\end{array}$

Therefore

Focus of parabola = $\left(a,0\right)$ = $\left(2,0\right)$

Vertex of parabola = $\left(0,0\right)$