Question

The graph of the equation $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ with $a>0,b>0$ is a hyperbola with (horizontal/vertical) transverse axis, vertices (…….. , ……….) and(………. , ………..) and foci$(0,\pm c)$, where c=. So the graph of $\frac{y^2}{4^2}-\frac{x^2}{3^2}=1$ is a hyperbola with vertices(………. , ……) and (………… , ………..) and foci (………. , ………..) and (……….. , ………..) .

Short answer

The vertices and foci of the hyperbola $\frac{y^2}{4^2}-\frac{x^2}{3^2}=1$ are $(0,\pm 4)$and $(0,\pm 5)$
respectively.

Step-by-step solution

Step 1. Given information.

The hyperbolic equation $\frac{y^2}{4^2}-\frac{x^2}{3^2}=1$.

Step 2. The vertices and foci of the hyperbola.

We will convert the given equation of the hyperbola in the standard form and then by comparing we will get the value of a,b.

Writing the given hyperbola equation in standard form we get,$\frac{y^2}{4^2}-\frac{x^2}{3^2}=1$

On comparing the giving hyperbola equation with the standard hyperbola equation we get,

$$\begin{gathered} a=4 \\ b=3 \end{gathered}$$

Now we have,

$\begin{array}{c}{c}^2=a^2+b^2\\ \Rightarrow c^2=4^2+3^2\\ \Rightarrow c^2=25\\ c=\pm 5\end{array}$

Hence the vertices are $\text{(0,4),(0,-4)}$
and the foci are $\text{(0,5),(0,-5)}$