Question

Identify the vertices, foci and equations for the asymptotes of the hyperbola below. Type coordinates with parentheses and separated by a comma like this (x,y). If a value is a non-integer then type is a decimal rounded to the nearest hundredth. \frac{(y-6)^2}{36}-\frac{(x+1)^2}{16}=1 The center is the point : AnswerThe vertex with a larger y value is the point :AnswerThe vertex with a smaller y value is the point :AnswerThe foci with a larger y value is the point :AnswerThe foci with a smaller y value is the point :AnswerOne of the asymptotes is the equation y=a(x+b)+c Where the value for a is: AnswerWhere the value for b is: AnswerWhere the value for c is: Answer

Answer 1

Given the equation of a hyperbola:

$\frac{(y-6)^2}{36}-\frac{(x+1)^2}{16}=1$

• You need to remember that the equation of a vertical hyperbola has this form:

$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1$

Where the center is:

$(h,k)$

In this case, you can identify that:

$\begin{gathered} h=-1 \\ k=6 \end{gathered}$

Therefore, its Center is at this point:

$(-1,6)$

• By definition, the Vertices of a vertical hyperbola can be found with:

$(h,k\pm a)$

In this case, you know that:

$a=\sqrt[]{36}=\pm6$

Therefore, you can determine that the Vertex with the larger y-value is:

$(-1,6+6)=(-1,12)$

And the Vertex with the smaller y-value is:

$(-1,6-6)=(-1,0)$

• By definition, the formula for calculating the distance from the Center of a hyperbola to the Foci is:

$c=\sqrt[]{a^2+b^2}$

You already know the value of "a", and you can determine that:

$b=\pm\sqrt[]{16}=\pm4$

Therefore, by substituting values into the formula and evaluating, you get:

$c=\sqrt[]{6^2+4^2}=2\sqrt[]{13}$

By definition, Focis of a vertical hyperbola have this form:

$(h,k\pm c)$

Hence, the Foci with a larger y-value is:

$(-1,6+2\sqrt[]{13})$

And the Foci with the smaller y-value is:

$(-1,6-2\sqrt[]{13})$

• According to the information provided in the exercise, one of the Asymptotes is the equation:

$y=a(x+b)+c$

By definition, the equation for the Asymptotes of a vertical hyperbola is:

$y=\pm\frac{a}{b}(x-h)+k$

Knowing all the values, you get:

$\begin{gathered} y=\pm\frac{6}{4}(x+1)+6 \\ \\ y=\pm\frac{3}{2}(x+1)+6 \end{gathered}$

Hence, the answers are:

• Center:

$(-1,6)$

• Vertex with a larger y-value:

$(-1,12)$

Vertex with a smaller y-value:

$(-1,0)$

Foci with a larger y-value:

$(-1,6+2\sqrt[]{13})$

Foci with a smaller y-value:

$(-1,6-2\sqrt[]{13})$

Values of "a", "b" and "c" of the equation of one of the asymptotes:

$\begin{gathered} a=\frac{3}{2} \\ \\ b=1 \\ \\ c=6 \end{gathered}$