EXPLANATION
Given the equation
(b)
[tex]9x^2-y^2=9[/tex]First, we need to find the vertex applying the following formula:
The vertices (h+a, k), (h-a,k) are the two bending points of the hyperbola with center (h,k) and semi-axis (a,b)
Calculate hyperbola properties:
Hyperbola standard equation:
[tex]\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1[/tex]Rewrite 9x^2 -y^2 = 9 in the form of a standard hyperbola equation:
Divide by coefficient of square terms: 9
[tex]x^2-\frac{16}{9}y^2=16[/tex]Divide by coefficient of square terms: 16
[tex]\frac{1}{16}x^2-\frac{1}{9}y^2=1[/tex]Refine:
[tex]\frac{x^2}{16}-\frac{y^2}{9}=1[/tex]Rewrite in standard form:
[tex]\frac{(x-0)^2}{4^2}-\frac{(y-0)^2}{3^2}=1[/tex]Therefore, hyperbola properties are:
(h,k) = (0,0) , a=4, b=3
Refine:
(4,0), (-4,0)
Now we need to compute the foci:
For a right-left faccing hyperbola, the foci are defined as (h+c,k), (h-c,k), where c= sqrt(a^2+b^2) is the distance from the center (h,k) to a focus.
Computing c (we have previously calculated a=4 and b=3):
[tex]c=\sqrt[]{4^2+3^2}=\sqrt[]{16+9}=\sqrt[]{25}=5[/tex]Refine:
Foci: (5,0), (-5,0)
Next, we need to find the asymptotes:
The asymptotes are the lines the hyperbola tends to at +- infinite
For right-left hyperbolas the asymptotes are:
[tex]y=\pm\frac{b}{a}(x-h)[/tex]Substituting terms:
[tex]y=\pm\frac{3}{4}(x-0)+0[/tex]Refine:
[tex]y=\frac{3x}{4},\text{ y=-}\frac{3x}{4}[/tex]Now, we need to find the vertices:
The endpoints are:
(4,0) , (-4,0)
Now, we need to find the center:
[tex]\text{center}=\text{ }\frac{x^2}{a^2}+\frac{y^2}{b^2}[/tex]