Answer
See graph
Work Step by Step
We are given the hyperbola:
$4x^2-25y^2-32x+164=0$
Bring the equation to the standard form:
$(4x^2-32x+64)-25y^2-64+164=0$
$4(x^2-8x+16)-25y^2+100=0$
$4(x-4)^2-25y^2+100=0$
$25y^2-4(x-4)^2=100$
$\dfrac{25y^2}{100}-\dfrac{4(x-4)^2}{100}=1$
$\dfrac{y^2}{4}-\dfrac{(x-4)^2}{25}=1$
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 $h,k,a,b,c$:
$h=4$
$k=0$
$a^2=4\Rightarrow a=\sqrt {4}=2$
$b^2=25\Rightarrow b=\sqrt {25}=5$
$c^2=a^2+b^2$
$c^2=4+25$
$c^2=29$
$c=\sqrt{29}$
The centre of the hyperbola is:
$(h,k)=(4,0)$
Determine the coordinates of the vertices:
$(h,k-a)=(4,0-2)=(4,-2)$
$(h,k+a)=(4,0+2)=(4,2)$
Determine the coordinates of the foci:
$(h,k-c)=(4,0-\sqrt{29})=(4,-\sqrt{29})$
$(h,k+c)=(4,0+\sqrt{29})=(4,\sqrt{29})$
Determine the asymptotes:
$y-k=\pm\dfrac{a}{b}(x-h)$
$y-0=\pm\dfrac{2}{5}(x-4)$
$y=\pm \dfrac{2}{5}(x-4)$
Graph the hyperbola:
