91Ó°ÊÓ

Focus is at $\left(-4,0\right)$and vertices are at$\left(-4,4\right)$and$\left(-4,2\right)$

The hyperbola with center $\left(h,k\right)$and transverse axis parallel to $y$-axis is

role="math" localid="1647054163294" $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$

where $b^2=c^2-a^2$

Focus at$\left(h,kÂ\pm c\right)$and vertices at$\left(h,kÂ\pm a\right)$

Since the one focus and vertices all lie on the line $x=-4$, the transverse axis is parallel to the $y$-axis.

So, $h=-4$

Since vertices are at localid="1647053566601" $\left(-4,4\right)$and $\left(-4,2\right)$,

$k+a=4$… (1)

$k-a=2$… (2)

Solving equations (1) and (2),

$k=3$

So, $a=1$

Also given that, the focus is at $\left(-4,0\right)$

So, $k-c=0$

Substituting value of k,

$c=3$

So, the center is at $\left(-4,3\right)$.

Since,$b^2=c^2-a^2$

$b^2=3^2-1^2$

So, $b^2=8$

Thus, the equation of the hyperbola will be

localid="1647054174030" $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$

Substituting all values,

localid="1647054256953" $\frac{{\left(y-3\right)}^2}{1}-\frac{{\left(x+4\right)}^2}{8}=1$

So,localid="1647054923963" ${\left(y-3\right)}^2-\frac{{\left(x+4\right)}^2}{8}=1$

The hyperbola is,

${\left(y-3\right)}^2-\frac{{\left(x+4\right)}^2}{8}=1$

Graph will be