Chapter 7, Problem 1RE

To determine

To identify: The given equation. If it is a parabola, give its vertex, focus and the directrix.

Answer to Problem 1RE

The vertex of the parabola $y^2=-16x$ is $\underset{\_}{\left(0,0\right)}$.

The focus of the parabola $y^2=-16x$ is $\underset{\_}{\left(-4,0\right)}$.

The directrix of the parabola $y^2=-16x$ is $\underset{\_}{x=4}$.

Explanation of Solution

Definitions used:

The standard equation of a parabola with vertex $\left(0,0\right)$ and the directrix $x=a$ is $y^2=-4ax$. The focus is $\left(-a,0\right)$ and the x-axis is the axis of symmetry. When $a>0$, the parabola opens to the right; when $a<0$, the parabola opens to the left.

Calculation:

Rewrite the equation $y^2=-16x$ as follows.

$y^2=-4\cdot 4\cdot x$

This equation is in the form of $y^2=-4ax$.

So, the given equation is parabola.

Therefore, the vertex of the parabola $y^2=-16x$ is $\underset{\_}{\left(0,0\right)}$.

Compare $y^2=-4\cdot 4\cdot x$ with $y^2=-4ax$.

Here, the value of a is 4.

Therefore, the focus of the parabola $y^2=-16x$ is $\underset{\_}{\left(-4,0\right)}$.

As $a=4$, the directrix is $x=4$.

Thus, the directrix of the parabola $y^2=-16x$ is $\underset{\_}{x=4}$.

Therefore, the vertex of the parabola $y^2=-16x$ is $\underset{\_}{\left(0,0\right)}$, the focus of the parabola is $\underset{\_}{\left(-4,0\right)}$ and the directrix of the parabola is $\underset{\_}{x=4}$.