What does the equation #(x-1)^2/4-(y+2)^2/9=1# tell me about its hyperbola?

2 Answers
Jul 23, 2018

Answer:

Please see the explanation below

Explanation:

The general equation of a hyperbola is

#(x-h)^2/a^2-(y-k)^2/b^2=1#

Here,

The equation is

#(x-1)^2/2^2-(y+2)^2/3^2=1#

#a=2#

#b=3#

#c=sqrt(a^2+b^2)=sqrt(4+9)=sqrt13#

The center is #C=(h,k)=(1,-2)#

The vertices are

#A=(h+a,k)=(3,-2)#

and

#A'=(h-a,k)=(-1,-2)#

The foci are

#F=(h+c,k)=(1+sqrt13,-2)#

and

#F'=(h-c,k)=(1-sqrt13,-2)#

The eccentricity is

#e=c/a=sqrt13/2#

graph{((x-1)^2/4-(y+2)^2/9-1)=0 [-14.24, 14.25, -7.12, 7.12]}

Answer:

See answer below

Explanation:

The given equation of hyperbola

#\frac{(x-1)^2}{4}-\frac{(y+2)^2}{9}=1#

#\frac{(x-1)^2}{2^2}-\frac{(y+2)^2}{3^2}=1#

The above equation is in standard form of hyperbola:

#(x-x_1)^2/a^2-(y-y_1)^2/b^2=1#

Which has

Eccentricity: #e=\sqrt{1+b^2/a^2}=\sqrt{1+9/4}=\sqrt13/2#

Center: #(x_1, y_1)\equiv(1, -2)#

Vertices: #(x_1\pm a, y_1)\equiv(1\pm2, -2)# &

#(x_1, y_1\pm b)\equiv(1, -2\pm 3)#

Asymptotes: #y-y_1=\pm b/a(x-x_1)#

#y+2=\pm3/2(x-1)#