Suggested languages for you:

Americas

Europe

Q 23.

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.

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 :

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

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

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

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

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

The graph of an equation ${y}^{2}=-8x$ will be

94% of StudySmarter users get better grades.

Sign up for free