Question

Find an equation of a hyperbola using foci and asymptotes

Answer 1

SOLUTION

Given the question in the image, the following are the solution steps to answer the question.

STEP 1: Write the given data values

$\begin{gathered} foci\Rightarrow(1,4-\sqrt{45}),(1,4+4+\sqrt{5}) \\ asymptotes\Rightarrow y=2x+2,y=-2x+6 \end{gathered}$

STEP 2: Write the equation

The equation of a hyperbola is given as:

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

where (h,k) is the center, a and b are the lengths of the semi-major and the semi-minor axes.

STEP 3: Explain the given data

$\begin{gathered} h=1,(k-4+3\sqrt{5})^2=a^2+b^2, \\ (k-3\sqrt{5}-4)^2=a^2+b^2 \\ \frac{b}{a}=2 \\ -2h+k=6 \end{gathered}$

STEP 4: Find the values of h,k,a and b

$\begin{gathered} h=1 \\ From: \\ -2h+k=2 \\ -2(1)+k=2 \\ -2+k=2 \\ k=2+2=4 \\ k=4 \\ \\ \text{To get b and a,} \\ \frac{b}{a}=2 \\ By\text{ calculation,} \\ a=3,b=6 \end{gathered}$

STEP 5: Get the equation of the hyperbola

$\begin{gathered} By\text{ substitution into the formula in step 1,} \\ \frac{(y-4)^2}{6^2}=\frac{(x-1)^2}{3^2}=1 \\ Vertex\text{ form}\Rightarrow\frac{(y-4)^2}{36}-\frac{(x-1)^2}{9}=1 \\ General\text{ form will be}\Rightarrow4x^2-8x-y^2+8y+24=0 \end{gathered}$

Hence, the equation of the hyperbola will be:

General form:

$\begin{equation*} 4x^2-8x-y^2+8y+24=0 \end{equation*}$

The vertex form is given as:

$\begin{equation*} \frac{(y-4)^2}{36}-\frac{(x-1)^2}{9}=1 \end{equation*}$