Chapter 9.1, Problem 53E

To determine

To find: The standard form of the equation of the parabola with vertex at origin and focus $\left(-\frac{1}{2},0\right)$ .

Answer to Problem 53E

The equation of the parabola with vertex at origin and focus $\left(-\frac{1}{2},0\right)$ is $y^2=-2x$ .

Explanation of Solution

Given information:

The vertex of the parabola is at origin and focus at $\left(-\frac{1}{2},0\right)$ .

Calculation:

The focus of the parabola lies on the $x$ -axis. So the parabola is horizontal axis parabola with vertex at origin and focus at $\left(-\frac{1}{2},0\right)$ .

The standard form of general equation of a horizontal axis parabola is ${\left(y-k\right)}^2=4p\left(x-h\right)$ .

The focus $\left(-\frac{1}{2},0\right)$ is $\frac{1}{2}$ units to the left of origin. So, the value of $p$ is $-\frac{1}{2}$ .

Substitute $0$ , $0$ for $h$ , $k$ and $-\frac{1}{2}$ for $p$ in the general equation.

  $\begin{array}{c}{\left(y-k\right)}^2=4p\left(x-h\right)\\ {\left(y-0\right)}^2=4\left(-\frac{1}{2}\right)\left(x-0\right)\\ y^2=-2x\end{array}$

Therefore, the equation of the parabola with vertex at origin and focus $\left(-\frac{1}{2},0\right)$ is $y^2=-2x$ .