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Expert-verified Found in: Page 643 ### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465 # Find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation by hand. Focus at $\left(-2,0\right)$ and directix of the line $x=2$.

The equation of a parabola is ${y}^{2}=-8x$. The points are $\left(-2,4\right)$ and $\left(-2,-4\right)$. The graph of an equation is : See the step by step solution

## Step 1. Given Information.

The given focus of an equation be $\left(-2,0\right)$ and an equation of its directix of a line be $x=2$.

## Step 2. Equation of a parabola.

The vertex of an equation be at $\left(0,0\right)$. The vertex is midway between the focus and the directix. Since, the focus is on the x-axis at $\left(-2,0\right)$, the equation of the parabola is f the form

role="math" localid="1646657517636" ${y}^{2}=4ax$ with $a=-2$.

i.e. ${y}^{2}=-8x$.

## Step 3. Latus rectum.

The points that determine the latus rectum by letting $x=-2$.

${y}^{2}=-8x\phantom{\rule{0ex}{0ex}}{y}^{2}=-8\left(-2\right)\phantom{\rule{0ex}{0ex}}{y}^{2}=16\phantom{\rule{0ex}{0ex}}y=±4$.

The points are $\left(-2,4\right)$ and $\left(-2,-4\right)$.

## Step 4. Graphing utility.

The graph of an equation ${y}^{2}=-8x$ will be  ### Want to see more solutions like these? 