Chapter 10, Problem 1RE

To determine

Tofind:the equation and if the equation is parabolic give its vertex, focus anddirectrix; if its ellipse gives its center, vertices andfoci; if its is a hyperbola, give its center, vertices, foci and asymptotes.

Answer to Problem 1RE

The vertex is $\left(0,0\right)$ ,Focus= $\left(\text{-a,0}\right)\text{=}\left(\text{-4,0}\right)$ and directrix is $\text{x=a}\Rightarrow \text{x=4}\text{.}$

Explanation of Solution

Given:

  ${\text{y}}^{\text{2}}\text{=-16x}$ .

Concept used:

The standard form is ${\left(\text{x-h}\right)}^{\text{2}}\text{=4p}\left(\text{y-k}\right)$ where the focus is $\left(\text{h,k+p}\right)$ and the directrix is $\text{y=k-p}$ . If the parabola is rotated so that its vertex is $\left(\text{h,k}\right)$ and its axis of symmetry is parallel to the x-axis, it has an equation of ${\left(\text{y-k}\right)}^{\text{2}}\text{=4p}\left(\text{x-h}\right)$ where the focus is $\left(\text{h+p,k}\right)$ and the directrix is $\text{x=h-p}$ .

Calculation:

  ${\text{y}}^{\text{2}}\text{=-16x}$ .

The standard form is ${\left(\text{x-h}\right)}^{\text{2}}\text{=4p}\left(\text{y-k}\right)$ where the focus is $\left(\text{h,k+p}\right)$ and the directrix is $\text{y=k-p}$ . If the parabola is rotated so that its vertex is $\left(\text{h,k}\right)$ and its axis of symmetry is parallel to the x-axis, it has an equation of ${\left(\text{y-k}\right)}^{\text{2}}\text{=4p}\left(\text{x-h}\right)$ where the focus is $\left(\text{h+p,k}\right)$ and the directrix is $\text{x=h-p}$ .

Here the equation is in the form of ${\text{y}}^{\text{2}}\text{=-4ax}$ where $\text{a=4}$ ,it’s a parabola with axis of symmetry $\text{x}$ and open left. The vertex is $\left(0,0\right)$ .

Focus= $\left(\text{-a,0}\right)\text{=}\left(\text{-4,0}\right)$ and directrix is $\text{x=a}\Rightarrow \text{x=4}\text{.}$

Hence,vertex is $\left(0,0\right)$ ,Focus= $\left(\text{-a,0}\right)\text{=}\left(\text{-4,0}\right)$ and directrix is $\text{x=a}\Rightarrow \text{x=4}\text{.}$