91Ó°ÊÓ

The hyperbola is given with the following equation

$\frac{y^2}{9}-\frac{x^2}{16}=1$

Since the variable $x$is being subtracted from the variable $y$it follow tha the transverse axis of the hyperbola is parallel to the $y-$axis

A hyperbola with its transverse axis parallel to the y-axis and its center in the point $(h,k)$has the following equation form:

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

By comparing, the center of the given hyperbola is in $(0,0)$

$a^2=9⇒a=3$

and

$b^2=16⇒b=4$

A hyperbola its transverse axis parallel to the $y-$axis and its origin. $(0,0)$

$(0,-a)$and $(0,a)$

Since you know $a=3$it follow that the vertices of the hyperbola are located in

$(0,-3)$and $(0,3)$

Again,

A hyperbola with its transverse axis parallel to the $y-$axis and its center in the origin, $(0,0)$

$(0,c)$and $(0,-c)$

Now,

$$\begin{gathered} c^2=3^2+4^2 \\ =9+16 \\ =25 \\ ⇒c=5 \end{gathered}$$

its foci are then

$(0,5)$and$(0,-5)$

You can make a drawing of the given hyperbola:

From the text of the exercise you know that the rear focus of the hyperbola is the same point as the focus of the parabola.

Since the focal length, the distance from the vertex of the parabola to the focus is $6$, it follow that the directrix of the parabola is $6$unit away from the vertex of the parabola

Then the vertex of the parabola is located in the point $(0,-11)$and the directrix is the line $y=-17$

A standard equation of a parabola with its of symmetry parallel to the $y-$axis is

${(x-h)}^2=4a(y-k)$

Since the focal length,$a$is $6$and the vertex of the parabola is in $(0,-11)$

$x^2=24(y+11)$