### About Conic Sections: The Hyperbola:

A hyperbola is the set of all points in a plane such that the absolute value of the difference between two fixed points is constant. The two fixed points are known as the foci.

Test Objectives
• Demonstrate the ability to sketch the graph of a hyperbola
• Demonstrate the ability to write the equation of a hyperbola
Conic Sections: The Hyperbola Practice Test:

#1:

Instructions: Sketch the graph of each, state the foci and vertices.

$$a)\hspace{.2em}\frac{(y + 1)^2}{16}- \frac{(x - 2)^2}{9}=1$$

$$b)\hspace{.2em}\frac{(x - 1)^2}{9}- \frac{(y + 2)^2}{4}=1$$

#2:

Instructions: Write each in standard form.

$$a)\hspace{.2em}-4x^2 + 9y^2 - 16x - 126y - 151=0$$

$$b)\hspace{.2em}-x^2 + y^2 + 20x + 8y - 120=0$$

#3:

Instructions: Write each in standard form.

$$a)\hspace{.2em}4x^2 - y^2 + 56x - 4y + 92=0$$

$$b)\hspace{.2em}-x^2 + 4y^2 + 2x - 64y + 239=0$$

#4:

Instructions: Write each in standard form.

$$a)\hspace{.2em}\text{vertices}: (7, 17), (7, 3)$$ $$\text{foci}: (7, 10 + \sqrt{149}), (7, 10 - \sqrt{149})$$

$$b)\hspace{.2em}\text{vertices}: (21, -9), (-3, -9)$$ $$\text{foci}: (24, -9), (-6, -9)$$

#5:

Instructions: Write each in standard form.

$$a)\hspace{.2em}\text{vertices}: (19, -4), (-5, -4)$$ $$\text{foci}: (20, -4), (-6, -4)$$

$$b)\hspace{.2em}\text{vertices}: (-4, 21), (-4, -9)$$ $$\text{foci}: (-4, 23), (-4, -11)$$

Written Solutions:

#1:

Solutions:

$$a)\hspace{.2em}\text{vertices}: (2, 3), (2, -5)$$ $$\text{foci}: (2, 4), (2, -6)$$

$$b)\hspace{.2em}\text{vertices}: (4, -2), (-2, -2)$$ $$\text{foci}: (1 + \sqrt{13}, -2), (1 - \sqrt{13}, -2)$$

#2:

Solutions:

$$a)\hspace{.2em}\frac{(y - 7)^2}{64}- \frac{(x + 2)^2}{144}=1$$

$$b)\hspace{.2em}\frac{(y + 4)^2}{36}- \frac{(x - 10)^2}{36}=1$$

#3:

Solutions:

$$a)\hspace{.2em}\frac{(x + 7)^2}{25}- \frac{(y + 2)^2}{100}=1$$

$$b)\hspace{.2em}\frac{(y - 8)^2}{4}- \frac{(x - 1)^2}{16}=1$$

#4:

Solutions:

$$a)\hspace{.2em}\frac{(y - 10)^2}{49}- \frac{(x - 7)^2}{100}=1$$

$$b)\hspace{.2em}\frac{(x - 9)^2}{144}- \frac{(y + 9)^2}{81}=1$$

#5:

Solutions:

$$a)\hspace{.2em}\frac{(x - 7)^2}{144}- \frac{(y + 4)^2}{25}=1$$

$$b)\hspace{.2em}\frac{(y - 6)^2}{225}- \frac{(x + 4)^2}{64}=1$$