Chapter 10.2, Problem 51AYU

In Problems $39-56$, find the vertex, focus, and directrix of each parabola. Graph the equation.

$y^2+2y-x=0$

To determine

The vertex, focus, directrix and graph of the parabola $y^2+2y-x=0$

Answer to Problem 51AYU

Solution:

The vertex of the parabola $y^2+2y-x=0$ is $\left(-1,-1\right)$

The focus of the parabola $y^2+2y-x=0$ is $\left(-\frac{3}{4},-1\right)$

The equation of the directrix is $x=-\frac{5}{4}$

The graph of the parabola $y^2+2y-x=0$ is

Explanation of Solution

Given information:

The equation of parabola $y^2+2y-x=0$

Explanation:

Consider the equation of parabola $y^2+2y-x=0$

To complete the square add $1$ to both sides

$\Rightarrow y^2+2y+1=x+1$

$\Rightarrow {\left(y+1\right)}^2=\left(x-\left(-1\right)\right)$

$\Rightarrow {\left(y-\left(-1\right)\right)}^2=\left(x-\left(-1\right)\right)$.

By comparing the function ${\left(y-\left(-1\right)\right)}^2=\left(x-0\right)$ with standard form of the parabola ${\left(y-k\right)}^2=4a\left(x-h\right)$, it gives $a=\frac{1}{4},h=-1,k=-1$

The vertex is $\left(h,k\right)$, the focus is $\left(h+a,k\right)$ and the equation of the directrix is $x=h-a$ of the parabola ${\left(y-k\right)}^2=4a\left(x-h\right)$

Then, $h+a=-1+\frac{1}{4}=-\frac{3}{4}$ and $h-a=-1-\frac{1}{4}=-\frac{5}{4}$

Hence, the vertex is $\left(-1,-1\right)$, the focus is $\left(-\frac{3}{4},-1\right)$ of the parabola $y^2+2y-x=0$

The equation of the directrix is $x=-\frac{5}{4}$ of the parabola $y^2+2y-x=0$

The axis of symmetry of the parabola ${\left(y-k\right)}^2=4a\left(x-h\right)$ is $y=k$, which is parallel to the $x-$ axis and the parabola opens to the right.

Hence, the axis of the symmetry of the parabola $y^2+2y-x=0$ is $y=-1$ which is parallel to the $x-$ axis and the parabola opens to the right.

Substitute $x=-\frac{3}{4}$ in the parabola $y^2+2y-x=0$,

$y^2+2y-\left(-\frac{3}{4}\right)=0$.

By using the quadratic formula $y=\frac{-2\pm \sqrt{4-4\left(\frac{3}{4}\right)}}{2}=\frac{-2\pm 1}{2}$

$\Rightarrow y=-\frac{3}{2}\text{or}y=-\frac{1}{2}$

The points of the latus rectum on the parabola are $\left(-\frac{3}{4},-\frac{3}{2}\right)$ and $\left(-\frac{3}{4},-\frac{1}{2}\right)$

By using all the information, the graph of the parabola $y^2+2y-x=0$ is as follows: