Question

I need this problem answered*answer the problem by using the answer options below in the picture*

Answer 1

Answer:

$\frac{(x+3)^2}{36^{}}-\frac{(y-2)^2}{16}=1$

Explanation:

The standard form of the equation of a hyperbola with center (h,k) and transverse axis parallel to the x-axis is:

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

Where:

• The length of the transverse axis = 2a

,

• The length of the conjugate axis = 2b

$\begin{gathered} \text{Centre,}(h,k)=(-3,2) \\ \text{The length of the transverse axis, 2a = 12 units}\implies a=\frac{12}{2}=6 \\ \text{The length of the conjugate axis, 2b = 8 units}\implies b=\frac{8}{2}=4 \end{gathered}$

Therefore. the equation of the hyperbola is:

$\begin{gathered} \frac{(x-(-3))^2}{6^2}-\frac{(y-2)^2}{4^2}=1 \\ \implies\frac{(x+3)^2}{36^{}}-\frac{(y-2)^2}{16}=1 \end{gathered}$