Question

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

Answer 1

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 2

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)`