Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Discover precise answers to your questions from a wide range of experts on our user-friendly Q&A platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
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]
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.