Answer :
To determine the vertices, foci, and asymptotes of the hyperbola defined by the equation [tex]\(9x^2 - 81y^2 = 729\)[/tex], let's follow these steps:
Step 1: Rewrite the equation in standard form
Given:
[tex]\[ 9x^2 - 81y^2 = 729 \][/tex]
First, divide every term by 729 to simplify it:
[tex]\[ \frac{9x^2}{729} - \frac{81y^2}{729} = 1 \][/tex]
[tex]\[ \frac{x^2}{81} - \frac{y^2}{9} = 1 \][/tex]
This is the standard form of a hyperbola:
[tex]\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \][/tex]
where [tex]\( a^2 = 81 \)[/tex] and [tex]\( b^2 = 9 \)[/tex].
Step 2: Determine the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]
[tex]\[ a = \sqrt{81} = 9 \][/tex]
[tex]\[ b = \sqrt{9} = 3 \][/tex]
Step 3: Find the vertices
The vertices of a hyperbola in this form ([tex]\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)[/tex]) are located at [tex]\( (\pm a, 0) \)[/tex].
Thus, the vertices are:
[tex]\[ (-9, 0) \text{ and } (9, 0) \][/tex]
Step 4: Find the foci
The foci of a hyperbola are given by [tex]\( (\pm c, 0) \)[/tex] where:
[tex]\[ c = \sqrt{a^2 + b^2} \][/tex]
[tex]\[ c = \sqrt{81 + 9} = \sqrt{90} = 3\sqrt{10} \][/tex]
Therefore, the foci are:
[tex]\[ (-3\sqrt{10}, 0) \text{ and } (3\sqrt{10}, 0) \][/tex]
Step 5: Find the equations of the asymptotes
The equations of the asymptotes for this hyperbola are:
[tex]\[ y = \pm \frac{b}{a} x \][/tex]
Using [tex]\( \frac{b}{a} = \frac{3}{9} = \frac{1}{3} \)[/tex], the asymptote equations are:
[tex]\[ y = \frac{1}{3}x \text{ and } y = -\frac{1}{3}x \][/tex]
Summary
- Vertices: [tex]\((-9, 0)\)[/tex] and [tex]\( (9, 0)\)[/tex]
- Foci: [tex]\((-3\sqrt{10}, 0)\)[/tex] and [tex]\( (3\sqrt{10}, 0)\)[/tex]
- Asymptotes: [tex]\( y = -\frac{1}{3}x \)[/tex] and [tex]\( y = \frac{1}{3}x \)[/tex]
Thus, we can fill in our solution as follows:
Vertices:
[tex]\( -9 \)[/tex] and [tex]\( 9 \)[/tex]
Foci:
[tex]\( -3\sqrt{10} \)[/tex] and [tex]\( 3\sqrt{10} \)[/tex]
Asymptotes:
[tex]\( y = -\frac{1}{3} \)[/tex]
[tex]\( \frac{1}{3}x \)[/tex] and [tex]\( y = \frac{1}{3}x \)[/tex]
So, the final answers are:
- Vertices: [tex]\( -9 \)[/tex] and [tex]\( 9 \)[/tex]
- Foci: [tex]\( -3\sqrt{10} \)[/tex] and [tex]\( 3\sqrt{10} \)[/tex]
- Asymptotes: [tex]\( y = -\frac{1}{3}x \)[/tex] and [tex]\( y = \frac{1}{3}x \)[/tex]
Step 1: Rewrite the equation in standard form
Given:
[tex]\[ 9x^2 - 81y^2 = 729 \][/tex]
First, divide every term by 729 to simplify it:
[tex]\[ \frac{9x^2}{729} - \frac{81y^2}{729} = 1 \][/tex]
[tex]\[ \frac{x^2}{81} - \frac{y^2}{9} = 1 \][/tex]
This is the standard form of a hyperbola:
[tex]\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \][/tex]
where [tex]\( a^2 = 81 \)[/tex] and [tex]\( b^2 = 9 \)[/tex].
Step 2: Determine the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]
[tex]\[ a = \sqrt{81} = 9 \][/tex]
[tex]\[ b = \sqrt{9} = 3 \][/tex]
Step 3: Find the vertices
The vertices of a hyperbola in this form ([tex]\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)[/tex]) are located at [tex]\( (\pm a, 0) \)[/tex].
Thus, the vertices are:
[tex]\[ (-9, 0) \text{ and } (9, 0) \][/tex]
Step 4: Find the foci
The foci of a hyperbola are given by [tex]\( (\pm c, 0) \)[/tex] where:
[tex]\[ c = \sqrt{a^2 + b^2} \][/tex]
[tex]\[ c = \sqrt{81 + 9} = \sqrt{90} = 3\sqrt{10} \][/tex]
Therefore, the foci are:
[tex]\[ (-3\sqrt{10}, 0) \text{ and } (3\sqrt{10}, 0) \][/tex]
Step 5: Find the equations of the asymptotes
The equations of the asymptotes for this hyperbola are:
[tex]\[ y = \pm \frac{b}{a} x \][/tex]
Using [tex]\( \frac{b}{a} = \frac{3}{9} = \frac{1}{3} \)[/tex], the asymptote equations are:
[tex]\[ y = \frac{1}{3}x \text{ and } y = -\frac{1}{3}x \][/tex]
Summary
- Vertices: [tex]\((-9, 0)\)[/tex] and [tex]\( (9, 0)\)[/tex]
- Foci: [tex]\((-3\sqrt{10}, 0)\)[/tex] and [tex]\( (3\sqrt{10}, 0)\)[/tex]
- Asymptotes: [tex]\( y = -\frac{1}{3}x \)[/tex] and [tex]\( y = \frac{1}{3}x \)[/tex]
Thus, we can fill in our solution as follows:
Vertices:
[tex]\( -9 \)[/tex] and [tex]\( 9 \)[/tex]
Foci:
[tex]\( -3\sqrt{10} \)[/tex] and [tex]\( 3\sqrt{10} \)[/tex]
Asymptotes:
[tex]\( y = -\frac{1}{3} \)[/tex]
[tex]\( \frac{1}{3}x \)[/tex] and [tex]\( y = \frac{1}{3}x \)[/tex]
So, the final answers are:
- Vertices: [tex]\( -9 \)[/tex] and [tex]\( 9 \)[/tex]
- Foci: [tex]\( -3\sqrt{10} \)[/tex] and [tex]\( 3\sqrt{10} \)[/tex]
- Asymptotes: [tex]\( y = -\frac{1}{3}x \)[/tex] and [tex]\( y = \frac{1}{3}x \)[/tex]
