Chapter 8.1, Problem 16E

Solution

To calculate : The equation in the standard form for the parabola that satisfies conditions focus $\left(-4,0\right)$ and directrix $x=4$ .

The equation in the standard form for the parabola that satisfies conditions focus $\left(-4,0\right)$ and directrix $x=4$ is $y^2=-16x$ .

Given information :

The parabola with focus $\left(-4,0\right)$ and directrix $x=4$ .

Formula used :

The equation for parabola for which focus $\left(p,0\right)$ is left to the vertex is $y^2=4px$ .

Calculation :

Consider the parabola with focus $\left(-4,0\right)$ and directrix $x=4$ ,

It can be seen that the focus and directrix are at equidistant from the vertex at $\left(0,0\right)$ .

As the focus is left to the vertex, the parabola opens to the left.

The equation of parabola is $y^2=4px$ .

As the focus is at $\left(-4,0\right)$ , then $p=-4$ . Substitute $-4$ for $p$ into the equation $y^2=4px$ .

  $\begin{array}{c}{y}^2=4\left(-4\right)x\\ =-16x\end{array}$

Thus, the equation in the standard form for the parabola that satisfies conditions focus $\left(-4,0\right)$ and directrix $x=4$ is $y^2=-16x$ .