### About Conic Sections: The Parabola:

A parabola is the set of all points that are equidistant from a fixed point called the focus and a fixed line called the directrix. Our goal for this section is to be able to find the vertex, focus, directrix, latus rectum, and equation when given information about the parabola.

Test Objectives
• Demonstrate the ability to find the vertex of a parabola
• Demonstrate the ability to find the focus of a parabola
• Demonstrate the ability to find the directrix of a parabola
• Demonstrate the ability to find the latus rectum of a parabola
• Demonstrate the ability to find the equation of a parabola
Conic Sections: The Parabola Practice Test:

#1:

Instructions: Find the vertex, focus, directrix, and endpoints of the latus rectum.

$$a)\hspace{.2em}y=\frac{1}{2}x^2 + 8x + 26$$

$$b)\hspace{.2em}x=\frac{1}{4}y^2 - \frac{7}{2}y + \frac{69}{4}$$

#2:

Instructions: Find the vertex, focus, directrix, and endpoints of the latus rectum.

$$a)\hspace{.2em} y = 2x^2 - 20x + 46$$

$$b)\hspace{.2em} y = -\frac{1}{4}x^2 - 3x - 15$$

#3:

Instructions: Find the vertex, focus, directrix, and endpoints of the latus rectum.

$$a)\hspace{.2em}2x^2 - 12x - y + 22 = 0$$

$$b)\hspace{.2em}{-}\frac{1}{8}y^2 + \frac{3}{2}y - x - \frac{17}{2}= 0$$

#4:

Instructions: Write an equation for the parabola.

$$a)\hspace{.2em}\text{Vertex:} \,(-10, -1)$$ $$\text{Focus:} \, \left(-\frac{15}{2}, -1\right)$$

$$b)\hspace{.2em}\text{Vertex:} \,(-3, 6)$$ $$\text{Focus:} \, \left(-\frac{25}{8}, 6\right)$$

#5:

Instructions: Write an equation for the parabola.

$$a)\hspace{.2em}\text{Vertex:} \, (-5, 10)$$ $$\text{Point:} \, \left(0, 9\right)$$ $$\text{Horizontal}$$

$$b)\hspace{.2em}\text{Vertex:} \, (-10, -4)$$ $$\text{Point:} \, \left(-8, -20\right)$$ $$\text{Vertical}$$

Written Solutions:

#1:

Solutions:

$$a)\hspace{.2em}$$ $$\text{vertex:} \, \left(-8, -6\right)$$ $$\text{focus:} \, \left(-8, -\frac{11}{2}\right)$$ $$\text{directrix:} \, y=-\frac{13}{2}$$ Endpoints of the latus rectum: $$\left(-9, -\frac{11}{2}\right), \left(-7, -\frac{11}{2}\right)$$

$$(x + 8)^2 = 2(y + 6)$$

$$b)\hspace{.2em}$$ $$\text{vertex:} \, \left(5, 7\right)$$ $$\text{focus:} \, \left(6, 7\right)$$ $$\text{directrix:} \, x=4$$ Endpoints of the latus rectum: $$\left(6, 5\right), \left(6, 9\right)$$

$$(y - 7)^2 = 4(x - 5)$$

#2:

Solutions:

$$a)\hspace{.2em}$$ $$\text{vertex:} \, \left(5, -4\right)$$ $$\text{focus:} \, \left(5, -\frac{31}{8}\right)$$ $$\text{directrix:} \, y = -\frac{33}{8}$$ Endpoints of the latus rectum: $$\left(\frac{19}{4}, -\frac{31}{8}\right), \left(\frac{21}{4}, -\frac{31}{8}\right)$$

$$(x - 5)^2 = \frac{1}{2}(y + 4)$$

$$b)\hspace{.2em}$$ $$\text{vertex:} \, \left(-6, -6\right)$$ $$\text{focus:} \, \left(-6, -7\right)$$ $$\text{directrix:} \, y = -5$$ Endpoints of the latus rectum: $$\left(-8, -7\right), \left(-4, -7\right)$$

$$(x + 6)^2 = -4(y + 6)$$

#3:

Solutions:

$$a)\hspace{.2em}$$ $$\text{vertex:} \, \left(3, 4\right)$$ $$\text{focus:} \, \left(3, \frac{33}{8}\right)$$ $$\text{directrix:} \, y = \frac{31}{8}$$ Endpoints of the latus rectum: $$\left(\frac{11}{4}, \frac{33}{8}\right), \left(\frac{13}{4}, \frac{33}{8}\right)$$

$$(x - 3)^2 = \frac{1}{2}(y - 4)$$

$$b)\hspace{.2em}$$ $$\text{vertex:} \, \left(-4, 6\right)$$ $$\text{focus:} \, \left(-6, 6\right)$$ $$\text{directrix:} \, x = -2$$ Endpoints of the latus rectum: $$\left(-6, 2\right), \left(-6, 10\right)$$

$$(y - 6)^2 = -8(x + 4)$$

#4:

Solutions:

$$a)\hspace{.2em}(y + 1)^2 = 10(x + 10)$$

$$b)\hspace{.2em}(y - 6)^2 = -\frac{1}{2}(x + 3)$$

#5:

Solutions:

$$a)\hspace{.2em} (y - 10)^2 = \frac{1}{5}(x + 5)$$

$$b)\hspace{.2em} (x + 10)^2 = -\frac{1}{4}(y + 4)$$