Question

9x^2- 4y^2- 36x + 8y - 4 = 01. does the hyperbola open horizontally or vertically?2. give the coordinates of the center 3. give the coordinates of the left vertex4. give the coordinates of the right vertex5. give the coordinates of the left focus (round to the hundredths)6. give the coordinates of the right vertex (round to the hundredths)

Answer 1

$\begin{gathered} (9x^2-36x)-(4y^2-8y)=4 \\ \text{Rewriting on standard form:} \\ \frac{(y-h)^2}{a^2}-\frac{(x-k)^2}{b^2}=1 \\ 9x^2-36x-4y^2-8y=4 \\ 9(x-4x+4)-4(y^2-2y+1)=4+36-4 \\ 9(x-2)^2-4(y-1)^2=36 \\ \text{Let's divide all the terms by 36:} \\ \frac{9(x-2)^2}{36}-\frac{4(y-1)^2}{36}=\frac{36}{36} \\ \frac{(x-2)^2}{4}-\frac{(y-1)^2}{9}=1 \end{gathered}$

If the x term is positive, that means the hyperbola opens horizontally.

To give the coordiantes of the verticesr:

$\begin{gathered} (h\pm a,\text{ k)} \\ \text{Left vertex:} \\ (2-2,\text{ 1)=}(0,\text{ 1)} \\ \text{Right vertex:} \\ (2+2,\text{ 1)}=(4,\text{ 1)} \end{gathered}$

For the coordinates of the foci or focus:

$\begin{gathered} (h\pm c,\text{ k)} \\ Where\text{ c=}\sqrt[]{a^2+b^2} \\ c=\sqrt[]{4+9}=\sqrt[\square]{13}=3.61 \\ \text{Left focus:} \\ (2-3.61,\text{ k)=(-1.61, 1)} \\ \text{Right focus:} \\ (2+3.61,\text{ k)=(5.61, 1)} \end{gathered}$

The center of the hyperbola is given by:

c(h,k)

c(2, 1)