Answer :
To find the area of the rectangle formed by the asymptotes of the hyperbola given by the equation [tex]\( 5x^2 - y^2 = 25 \)[/tex], we need to convert the equation into its standard form.
### Step-by-step solution:
1. Rewrite the Hyperbola in Standard Form:
The given equation of the hyperbola is:
[tex]\[ 5x^2 - y^2 = 25 \][/tex]
To convert this to the standard form of a hyperbola, we divide the entire equation by 25:
[tex]\[ \frac{5x^2}{25} - \frac{y^2}{25} = \frac{25}{25} \][/tex]
Simplify the equation:
[tex]\[ \frac{x^2}{5} - \frac{y^2}{25} = 1 \][/tex]
Now we have the standard form:
[tex]\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \quad \text{with} \quad a^2 = 5 \quad \text{and} \quad b^2 = 25 \][/tex]
2. Find [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
- [tex]\(a^2 = 5\)[/tex] implies [tex]\(a = \sqrt{5}\)[/tex]
- [tex]\(b^2 = 25\)[/tex] implies [tex]\(b = 5\)[/tex]
3. Understand the Asymptotes:
The asymptotes of the hyperbola are given by the equations:
[tex]\[ y = \pm \frac{b}{a} x \][/tex]
Therefore, the rectangle formed by the asymptotes extends to [tex]\(a\)[/tex] units along the [tex]\(x\)[/tex]-axis and [tex]\(b\)[/tex] units along the [tex]\(y\)[/tex]-axis.
4. Area of the Rectangle:
The length and breadth of the rectangle are [tex]\(2a\)[/tex] and [tex]\(2b\)[/tex] respectively because the asymptotes extend on both sides of the origin.
So, the area of the rectangle formed by the asymptotes is:
[tex]\[ 2a \times 2b \][/tex]
Substitute [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
[tex]\[ 2\sqrt{5} \times 2 \times 5 = 4\sqrt{5} \times 5 = 20\sqrt{5} \][/tex]
So, the area of the asymptote rectangle is [tex]\( 20 \sqrt{5} \)[/tex] square units.
Correct answer: A. [tex]\( 20 \sqrt{5} \)[/tex] square units
### Step-by-step solution:
1. Rewrite the Hyperbola in Standard Form:
The given equation of the hyperbola is:
[tex]\[ 5x^2 - y^2 = 25 \][/tex]
To convert this to the standard form of a hyperbola, we divide the entire equation by 25:
[tex]\[ \frac{5x^2}{25} - \frac{y^2}{25} = \frac{25}{25} \][/tex]
Simplify the equation:
[tex]\[ \frac{x^2}{5} - \frac{y^2}{25} = 1 \][/tex]
Now we have the standard form:
[tex]\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \quad \text{with} \quad a^2 = 5 \quad \text{and} \quad b^2 = 25 \][/tex]
2. Find [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
- [tex]\(a^2 = 5\)[/tex] implies [tex]\(a = \sqrt{5}\)[/tex]
- [tex]\(b^2 = 25\)[/tex] implies [tex]\(b = 5\)[/tex]
3. Understand the Asymptotes:
The asymptotes of the hyperbola are given by the equations:
[tex]\[ y = \pm \frac{b}{a} x \][/tex]
Therefore, the rectangle formed by the asymptotes extends to [tex]\(a\)[/tex] units along the [tex]\(x\)[/tex]-axis and [tex]\(b\)[/tex] units along the [tex]\(y\)[/tex]-axis.
4. Area of the Rectangle:
The length and breadth of the rectangle are [tex]\(2a\)[/tex] and [tex]\(2b\)[/tex] respectively because the asymptotes extend on both sides of the origin.
So, the area of the rectangle formed by the asymptotes is:
[tex]\[ 2a \times 2b \][/tex]
Substitute [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
[tex]\[ 2\sqrt{5} \times 2 \times 5 = 4\sqrt{5} \times 5 = 20\sqrt{5} \][/tex]
So, the area of the asymptote rectangle is [tex]\( 20 \sqrt{5} \)[/tex] square units.
Correct answer: A. [tex]\( 20 \sqrt{5} \)[/tex] square units
