Question

Find an equation of a hyperbola using foci and asymptotes

Answer 1

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-√(45)),(1,4+4+√(5)) \\ asymptotes\Rightarrow y=2x+2,y=-2x+6 \end{gathered}`

STEP 2: Write the equation

The equation of a hyperbola is given as:

`((y-k)^2)/(b^2)-((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√(5))^2=a^2+b^2, \\ (k-3√(5)-4)^2=a^2+b^2 \\ (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,} \\ (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,} \\ ((y-4)^2)/(6^2)=((x-1)^2)/(3^2)=1 \\ Vertex\text{ form}\Rightarrow((y-4)^2)/(36)-((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*} ((y-4)^2)/(36)-((x-1)^2)/(9)=1 \end{equation*}`