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. -4x^2+24x+16y^2-128y+156=0 The center is the point :

Answer 1

ANSWER:

The center is:

$(3,4)$

Vertex with larger y-value:

$(3,6)$

Vertex with smaller y-value:

$(3,2)$

Foci with larger y-value:

$(3,8)$

Foci with smaller y-value:

$(3,0)$

Equation of an asymptote:

$y=0.5(x-3)+4$

Where

$\begin{gathered} a=0.5 \\ b=3 \\ c=4 \end{gathered}$

EXPLANATION:

We have to take this equation into the general form of an hyperbola:

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

Where (h,k) is the center of the hyperbola.

We also know the vertices are:

$\begin{gathered} (h,k+b) \\ (h,k-b) \end{gathered}$

The foci are:

$\begin{gathered} (h,k+2b) \\ (h,k-2b) \end{gathered}$

And that the asymptotes are given by the expression:

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

Let's manipulate the equation:

$\begin{gathered} -4x^2+24x+16y^2-128y+156=0 \\ \rightarrow16y^2-128y-4x^2+24x+156=0 \\ \rightarrow(16y^2-128y-4x^2+24x+156)\div4=0\div4 \\ \rightarrow4y^2-32y-x^2+6x+39=0 \\ \rightarrow4(y-4)^2-64-(x-3)^2+9+39=0 \\ \rightarrow4(y-4)^2-(x-3)^2-16=0 \\ \rightarrow4(y-4)^2-(x-3)^2=16 \\ \\ \rightarrow\frac{\mleft(y-4\mright)^{}_{}^2}{4}-\frac{(x-3)^2}{16}=1 \\ \\ \Rightarrow\frac{(y-4)^2_{}}{2^2}-\frac{(x-3)^2}{4^2}=1 \end{gathered}$

From this general equation, we can conclude that the center is:

$(3.4)$

Now, the vertices are:

$\begin{gathered} (3,4+2)\rightarrow(3,6) \\ (3,4-2)\rightarrow(3,2) \end{gathered}$

The foci are:

$\begin{gathered} (3,4+4)\rightarrow(3,8) \\ (3,4-4)\rightarrow(3,0) \end{gathered}$

And the equation of the asympotes are:

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

One of this asymptotes is:

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