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Q 23.

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