Extension Test: Conics and Foci Solutions

1. $\frac{y^2}{9}–\frac{x^2}{16}=1$

Orientation of hyperbola: vertical

Center: $(0, 0)$

Length of a and b:

$$\begin{gathered} a^2=9, a=3 \\ {b}^2=16, b=4 \end{gathered}$$

Focal length (c):

$\begin{array}{rcl}{a}^2+b^2& =& c^2\\ 3^2+4^2& =& c^2\\ c^2& =& 25\\ c& =& 5\end{array}$

Coordinates of foci:

$$\begin{gathered} \left(h, k\pm c\right) \\ \left(0, 0+5\right)=\left(0, 5\right) \\ \left(0, 0–5\right)=\left(0, –5\right) \end{gathered}$$

2. $\frac{{\left(x–2\right)}^2}{169}+\frac{{\left(y+1\right)}^2}{25}=1$

Orientation of ellipse: horizontal

Center: $\left(2, –1\right)$

Length of a and b:

$$\begin{gathered} a^2=169, a=13 \\ {b}^2=25, b=5 \end{gathered}$$

Focal length (c):

$\begin{array}{rcl}{a}^2–b^2& =& c^2\\ 169–25& =& 144\\ c^2& =& 144\\ c& =& 12\end{array}$

Coordinates of foci:

$$\begin{gathered} \left(h\pm c, k\right) \\ \left(2+12, –1\right)=\left(14, –1\right) \\ \left(2–12, –1\right)=\left(–10, –1\right) \end{gathered}$$

3. An ellipse has a vertical major axis that is 10 units long and focal points at $\left(2, 1\right) \mathrm{and} \left(2, 9\right)$.

Center:
$\left(\frac{2+2}{2}, \frac{1+9 }{2}\right)=\left(2, 5\right)$

Vertical major axis (a):

$$\begin{gathered} 2a=10 \\ a=5 \end{gathered}$$

Focal length (c):

$$\begin{gathered} \left(2, 5\right), \left(2, 9\right) \\ c=\sqrt{{\left(2–2\right)}^2+{\left(9–5\right)}^2}=\sqrt{16} \\ c=4 \end{gathered}$$

Find b:

$\begin{array}{rcl}{a}^2–b^2& =& c^2\\ 5^2–b^2& =& 4^2\\ 25–b^2& =& 16\\ –b^2& =& –9\\ b& =& 3\end{array}$

Equation: $\frac{{\left(x–2\right)}^2}{9}+\frac{{\left(y–5\right)}^2}{25}=1$

4. Determine the equation of the hyperbola in standard form with vertices at $\left(–2, 3\right) \mathrm{and} \left(4, 3\right)$ and foci at $\left(–4, 3\right) \mathrm{and} \left(6, 3\right)$.

Center: $\left(\frac{–2+4}{2}, \frac{3+3 }{2}\right)=\left(1, 3\right)$

Focal length (c):

$$\begin{gathered} \left(1, 3\right), \left(6, 3\right) \\ c=\sqrt{{\left(6–1\right)}^2+{\left(3–3\right)}^2}=\sqrt{25} \\ c=5 \end{gathered}$$

Find a:

$$\begin{gathered} \left(–2, 3\right), \left(1, 3\right) \\ a=\sqrt{{\left(1–\left(–2\right)\right)}^2+{\left(3–3\right)}^2}=\sqrt{9} \\ a=3 \end{gathered}$$

Length of b:

$$\begin{gathered} a^2+b^2=c^2 \\ {3}^2+b^2=5^2 \\ 9+b^2=25 \\ {b}^2=16 \\ b=4 \end{gathered}$$

Equation: $\frac{{\left(x–1\right)}^2}{9}–\frac{{\left(y–3\right)}^2}{16}=1$