Question

find the equation of the hyperbola with the following properties. express your answer in standard form. vertices at (-3,1) and (1,1) asymptotes of ( y - 1=pm\frac{7}{2}(x + 1) )

Explanation:

Step1: Find the center of the hyperbola

The center \((h,k)\) of a hyperbola is the mid - point of the vertices.
If the vertices are \((x_1,y_1)=(-3,1)\) and \((x_2,y_2)=(1,1)\), then \(h=\frac{-3 + 1}{2}=-1\), \(k = 1\).

Step2: Find the value of \(a\)

The distance between the center \((h,k)=(-1,1)\) and a vertex \((x,y)\) is \(a\). Using the distance formula \(d=\sqrt{(x - h)^2+(y - k)^2}\) or for a horizontal hyperbola (since the \(y\) - coordinates of the vertices are the same), \(a=\vert x - h\vert\).
Take the vertex \((1,1)\), then \(a=\vert1-(-1)\vert = 2\).

Step3: Find the value of \(b\)

The equation of the asymptotes for a horizontal hyperbola \(\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1\) is \(y - k=\pm\frac{b}{a}(x - h)\).
We are given \(y - 1=\pm\frac{7}{2}(x + 1)\), and \(a = 2\). Since \(\frac{b}{a}=\frac{7}{2}\) and \(a = 2\), then \(b = 7\).

Step4: Write the equation of the hyperbola

Substitute \(h=-1\), \(k = 1\), \(a = 2\), and \(b = 7\) into the standard form of a horizontal hyperbola \(\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1\).
We get \(\frac{(x+1)^2}{4}-\frac{(y - 1)^2}{49}=1\)

Answer:

\(\frac{(x + 1)^2}{4}-\frac{(y - 1)^2}{49}=1\)