Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Discover comprehensive solutions to your questions from a wide network of experts on our user-friendly platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A 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]
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.
