Question

question 19 of 20
what is the length of the transverse axis of the conic section shown below?
\\(\frac{(y + 2)^2}{16}-\frac{(x - 3)^2}{9}=1\\)

Explanation:

Step1: Identify the form of hyperbola

The given equation $\frac{(y + 2)^2}{16}-\frac{(x - 3)^2}{9}=1$ is of the form $\frac{(y - k)^2}{a^2}-\frac{(x - h)^2}{b^2}=1$, which is a vertical - oriented hyperbola.

Step2: Determine the value of $a$

For a hyperbola of the form $\frac{(y - k)^2}{a^2}-\frac{(x - h)^2}{b^2}=1$, we have $a^2 = 16$, so $a=\sqrt{16}=4$.

Step3: Calculate the length of the transverse axis

The length of the transverse axis of a hyperbola of the form $\frac{(y - k)^2}{a^2}-\frac{(x - h)^2}{b^2}=1$ is $2a$. Substituting $a = 4$, we get $2a=8$.

Answer:

8