Answer
See graph
Work Step by Step
We are given the hyperbola:
$9x^2-36x-16y^2-64y+116=0$
Bring the equation to the standard form:
$(9x^2-36x+36)-(16y^2+64y+64)-36+64+116=0$
$9(x^2-4x+4)-16(y^2+4y+4)+144=0$
$9(x-2)^2-16(y+2)^2+144=0$
$16(y+2)^2-9(x-2)^2=144$
$\dfrac{16(y+2)^2}{144}-\dfrac{9(x-2)^2}{144}=1$
$\dfrac{(y+2)^2}{9}-\dfrac{(x-2)^2}{16}=1$
The equation is in the form:
$\dfrac{(y-k)^2}{a^2}-\dfrac{(x-h)^2}{b^2}=1$
The transverse axis is parallel to the $y$-axis.
Determine $h,k,a,b,c$:
$h=2$
$k=-2$
$a^2=9\Rightarrow a=\sqrt {9}=3$
$b^2=16\Rightarrow b=\sqrt {16}=4$
$c^2=a^2+b^2$
$c^2=9+16$
$c^2=25$
$c=\sqrt{25}$
$c=5$
The centre of the hyperbola is:
$(h,k)=(2,-2)$
Determine the coordinates of the vertices:
$(h,k-a)=(2,-2-3)=(2,-5)$
$(h,k+a)=(2,-2+3)=(2,1)$
Determine the coordinates of the foci:
$(h,k-c)=(2,-2-5)=(2,-7)$
$(h,k+c)=(2,-2+5)=(2,3)$
Determine the asymptotes:
$y-k=\pm\dfrac{a}{b}(x-h)$
$y+2=\pm\dfrac{3}{4}(x-2)$
Graph the hyperbola:
