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

2 Answers
Narad T.
Jul 23, 2018

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]}

Harish Chandra Rajpoot
Jul 23, 2018

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)

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