Chapter 10.2, Problem 40AYU

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

$x^2=-4y$

To determine

The vertex, focus, directrix and graph of the parabola $x^2=-4y$.

Answer to Problem 40AYU

Solution:

The vertex of the parabola $x^2=-4y$ is $\left(0,0\right)$.

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

The equation of the directrix is $y=1$ .

The graph of the parabola $x^2=-4y$ is

Explanation of Solution

Given information:

The Equation of parabola $x^2=-4y$

Explanation:

By comparing the equation of parabola $x^2=-4y$ with standard form of the parabola $x^2=-4ay$, it gives $a=1$.

The vertex is $\left(0,0\right)$, the focus is $\left(0,-a\right)$ and the equation of the directrix is $y=a$ of the parabola $x^2=-4ay$.

Hence, the vertex is $\left(0,0\right)$, the focus is $\left(0,-1\right)$ of the parabola $x^2=-4y$.

The equation of the directrix is $y=1$ of the parabola $x^2=-4y$.

The axis of symmetry of the parabola $x^2=-4ay$ is $y-$ axis and the parabola opens down.

Hence the axis of symmetry of the parabola $x^2=-4y$ is $y-$ axis and the parabola opens down.

Substitute $y=-1$ in the parabola $x^2=-4y$,

$x^2=-4y$

$\Rightarrow x^2=-4\left(-1\right)$

$\Rightarrow x=\pm 2$.

The points of the latus rectum on the parabola are $\left(-2,-1\right)$ and $\left(2,-1\right)$.

By using allinformation, the graph of the parabola $x^2=-4y$ is as follows: