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Q 26.

Expert-verifiedFound in: Page 643

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.

line of directix $x=-\frac{1}{2}$ and vertex at the point $\left(0,0\right)$.

The equation of a parabola is ${y}^{2}=2x$. The two points are $\left(\frac{1}{2},1\right)$ and $\left(\frac{1}{2},-1\right)$.

The graph of an equation is

The given vertex is at the point $\left(0,0\right)$ and the line of directix is $x=-\frac{1}{2}$.

A parabola whose focus is at $\left(\frac{1}{2},0\right)$ and whose directix is the line $x=-\frac{1}{2}$ has its vertex at $\left(0,0\right)$.The equation of a parabola is of the form as

${y}^{2}=4ax\phantom{\rule{0ex}{0ex}}{y}^{2}=2x$

where $a=\frac{1}{2}$.

The two points that determines the latus rectum by letting $x=\frac{1}{2}$.Then,

${y}^{2}=2x\phantom{\rule{0ex}{0ex}}{y}^{2}=2\left(\frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}{y}^{2}=1\phantom{\rule{0ex}{0ex}}y=\pm 1$.

The points are $\left(\frac{1}{2},1\right)$ and $\left(\frac{1}{2},-1\right)$.

The graph of a parabola is

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