Given equation of hyperbola:
\frac{(y-4)^2}{4^2}-\frac{(x+2)^2}{3^2}=1(y−4)242−(x+2)232=1
Comparing above equation with standard form of vertical hyperbola: \frac{Y^2}{a^2}-\frac{X^2}{b^2}=1Y2a2−X2b2=1 we get
X=x+2, Y=y-4, a=4, b=3X=x+2,Y=y−4,a=4,b=3
\text{eccentricity}, e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{3^2}{4^2}}=5/4eccentricity,e=√1+b2a2=√1+3242=54
**Center of hyperbola: **
X=0, Y=0X=0,Y=0
x+2=0, y-4=0x+2=0,y−4=0
x=-2, y=4x=−2,y=4
\text{Center}\equiv(-2, 4)Center≡(−2,4)
Now, the vertices of hyper bola
X=0, Y=\pmaX=0,Y=±a
x+2=0, y-4=\pm4x+2=0,y−4=±4
x=-2, y=4\pm4x=−2,y=4±4
\text {Vertices} (-2, 8)\ & \ (-2, 0)
Now, the focii of hyper bola
X=0, Y=\pm ae
x+2=0, y-4=\pm 4\cdot 5/4
x=-2, y=4\pm5
\text {focii}, (-2, 9)\ &\ (-2, -1)
Asymptotes of hyperbola
Y=\pma/bX
y-4=\pm4/3(x+2)
y=\pm4/3(x+2)+4
