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Q 30.

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.

Vertex is at $\left(4,-2\right)$ and focus is at $\left(6,-2\right)$.

The equation of a parabola is

${\left(y+2\right)}^{2}=8\left(x-4\right)$.

The points are $\left(6,2\right);\left(6,-6\right)$.

The graph of an equation is

The given vertex is at $\left(4,-2\right)$ and focus is at $\left(6,-2\right)$.

The vertex is at $\left(4,-2\right)$ and focus $\left(6,-2\right)$ both lie on the horizontal line $y=-2$ (the axis of symmetry). The distance a from the vertex to the focus is $a=-2$.

The parabola opens to the right. The form of a equation is

${\left(y-k\right)}^{2}=-4a\left(x-h\right)$.

where $\left(h,k\right)=\left(4,-2\right)$ and $a=-2$. Therefore, the equation is

${\left(y+2\right)}^{2}=8\left(x-4\right)$.

The two points that determines the latus rectum by letting $x=6$, so that

${\left(y+2\right)}^{2}=8\left(x-4\right)\phantom{\rule{0ex}{0ex}}{\left(y+2\right)}^{2}=8\left(2\right)\phantom{\rule{0ex}{0ex}}{\left(y+2\right)}^{2}=16\phantom{\rule{0ex}{0ex}}y+2=\pm 4\phantom{\rule{0ex}{0ex}}y+2=4,y+2=-4\phantom{\rule{0ex}{0ex}}y=2,-6.$

The points are $\left(6,2\right)$ and $\left(6,-6\right)$.

The graph of a parabola is

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