Chapter 9, Problem 17PT

To determine

To find: The has vertices at $\left(-3,-3\right)and\left(5,-3\right)$ and axis of length $6$ units.

Answer to Problem 17PT

The option B is correct option.

Explanation of Solution

Given information:

The given expression is.

  $\begin{array}{l}A\frac{{\left(y-1\right)}^2}{16}-\frac{{\left(x+3\right)}^2}{9}=1\\ B\frac{{\left(x-1\right)}^2}{16}-\frac{{\left(y+3\right)}^2}{9}=1\\ C\frac{{\left(y+1\right)}^2}{16}-\frac{{\left(x-3\right)}^2}{9}=1\\ D\frac{{\left(x+1\right)}^2}{16}-\frac{{\left(y-3\right)}^2}{9}=1\end{array}$ .

Calculation:

Calculate the expression.

The midpoint of the vertices will be center of the hyperbola.

  $\begin{array}{c}\left(h,k\right)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\ =\left(\frac{\left(-3\right)+5}{2},\frac{\left(-3\right)+\left(-3\right)}{2}\right)\end{array}$ .

The distance between the vertices will be equal to $2a$ .

  $\begin{array}{c}2a=5-\left(-3\right)\\ =8\\ a=4\end{array}$

Conjugate axis is of length 6 units.

  $\begin{array}{c}2b=6\\ b=3\end{array}$

Use the formula.

  $\begin{array}{l}\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y+k\right)}^2}{b^2}=1\\ \end{array}$

Substitute the value $\left(h,k\right)=\left(1,-3\right),a=4andb=3$

  $\begin{array}{l}\frac{{\left(x-1\right)}^2}{4^2}-\frac{{\left(x+\left(-3\right)\right)}^2}{3^2}=1\\ \frac{{\left(y-1\right)}^2}{16}-\frac{{\left(x+3\right)}^2}{9}=1\end{array}$

Therefore the option B is correct option.