Answer
See graph
Work Step by Step
We are given the hyperbola:
$4x^2-9y^2-16x+54y-101=0$
Bring the equation to the standard form:
$(4x^2-16x+16)-(9y^2-54y+81)-16+81-101=0$
$4(x^2-4x+4)-9(y^2-6y+9)-36=0$
$4(x-2)^2-9(y-3)^2=36$
$\dfrac{4(x-2)^2}{36}-\dfrac{9(y-3)^2}{36}=1$
$\dfrac{(x-2)^2}{9}-\dfrac{(y-3)^2}{4}=1$
The equation is in the standard form:
$\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$
The transverse axis is parallel to the $x$-axis.
Determine $h,k,a,b,c$:
$h=2$
$k=3$
$a^2=9\Rightarrow a=\sqrt 9=3$
$b^2=4\Rightarrow b=\sqrt 4=2$
$c^2=a^2+b^2$
$c^2=9+4$
$c^2=13$
$c=\sqrt{13}$
The centre of the hyperbola is:
$(h,k)=(2,3)$
Determine the coordinates of the vertices:
$(h-a,k)=(2-3,3)=(-1,3)$
$(h+a,k)=(2+3,3)=(5,3)$
Determine the coordinates of the foci:
$(h-c,k)=(2-\sqrt{13},3)$
$(h+c,k)=(2+\sqrt{13},3)$
Determine the asymptotes:
$y-k=\pm\dfrac{b}{a}(x-h)$
$y-3=\pm\dfrac{2}{3}(x+2)$
Graph the hyperbola:
