Question

I need this problem from my prep guide to be answered.

Answer 1

Given,

The center of the hyperbola is (-3, 2).

The length of the conjugate axis is 8 units.

The length of transverse axis is 12 units.

The center of the hyperbola is ,

`\begin{gathered} (h,k)=(-3,2) \\ So,\text{ h = -3 and k =2} \end{gathered}`

As, the transverse axis is parallel to the x-axis then hyperbola must be opens on left and right.

The length of conjugate axis is calculated as:

`\begin{gathered} \text{Length of conjugate axis = 2b} \\ 8=2b \\ b=4 \end{gathered}`

The length of transverse axis is calculated as:

`\begin{gathered} \text{Length of transverse axis = 2a} \\ 12=2a \\ a=6 \end{gathered}`

The standard equation of hyperbola is,

`((x-h)^2)/(a^2)-((y-b)^2)/(b^2)=1`

Substituting the values then,

`\begin{gathered} ((x-(-3))^2)/(6^2)-((y-2)^2)/(4^2)=0 \\ \frac{(x+3)^2}{36^{}}-\frac{(y-2)^2}{16^{}}=0 \end{gathered}`

Hence, the equation of hyperbola is (x+3)^2/36-(y-2)^2/16=0