Answer :
To solve the problem of determining the boundaries of the light emitted by the lamp, we need to find the equations of the asymptotes of the hyperbola given by [tex]\(16x^2 - 9y^2 + 576 = 0\)[/tex].
Here’s a detailed step-by-step method to find the asymptotes:
1. Rearrange the given equation into standard form:
Start by moving the constant term to the other side:
[tex]\[ 16x^2 - 9y^2 = -576 \][/tex]
2. Divide both sides by [tex]\(-576\)[/tex] to simplify:
We want to rewrite the equation in the form [tex]\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)[/tex]:
[tex]\[ \frac{16x^2}{-576} - \frac{9y^2}{-576} = \frac{-576}{-576} \][/tex]
Simplify the fractions:
[tex]\[ \frac{x^2}{-36} - \frac{y^2}{-64} = 1 \][/tex]
Multiply the entire equation by [tex]\(-1\)[/tex] to match the standard form for a hyperbola:
[tex]\[ \frac{x^2}{36} - \frac{y^2}{64} = -1 \][/tex]
3. Identify [tex]\(a^2\)[/tex] and [tex]\(b^2\)[/tex]:
The equation is now:
[tex]\[ \frac{x^2}{36} - \frac{y^2}{64} = -1 \][/tex]
Here, [tex]\( a^2 = 36 \)[/tex] and [tex]\( b^2 = 64 \)[/tex]. Therefore, [tex]\( a = \sqrt{36} = 6 \)[/tex] and [tex]\( b = \sqrt{64} = 8 \)[/tex].
4. Find the slopes of the asymptotes:
The asymptotes of the hyperbola [tex]\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = -1\)[/tex] are given by the equations:
[tex]\[ y = \pm \frac{b}{a} x \][/tex]
Calculate [tex]\(\frac{b}{a}\)[/tex]:
[tex]\[ \frac{b}{a} = \frac{8}{6} = \frac{4}{3} \][/tex]
5. Write the equations of the asymptotes:
Therefore, the equations of the asymptotes are:
[tex]\[ y = \frac{4}{3} x \quad \text{and} \quad y = -\frac{4}{3} x \][/tex]
So, the correct answer is:
[tex]\[ \boxed{y = \frac{4}{3} x \text{ and } y = -\frac{4}{3} x} \][/tex]
Here’s a detailed step-by-step method to find the asymptotes:
1. Rearrange the given equation into standard form:
Start by moving the constant term to the other side:
[tex]\[ 16x^2 - 9y^2 = -576 \][/tex]
2. Divide both sides by [tex]\(-576\)[/tex] to simplify:
We want to rewrite the equation in the form [tex]\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)[/tex]:
[tex]\[ \frac{16x^2}{-576} - \frac{9y^2}{-576} = \frac{-576}{-576} \][/tex]
Simplify the fractions:
[tex]\[ \frac{x^2}{-36} - \frac{y^2}{-64} = 1 \][/tex]
Multiply the entire equation by [tex]\(-1\)[/tex] to match the standard form for a hyperbola:
[tex]\[ \frac{x^2}{36} - \frac{y^2}{64} = -1 \][/tex]
3. Identify [tex]\(a^2\)[/tex] and [tex]\(b^2\)[/tex]:
The equation is now:
[tex]\[ \frac{x^2}{36} - \frac{y^2}{64} = -1 \][/tex]
Here, [tex]\( a^2 = 36 \)[/tex] and [tex]\( b^2 = 64 \)[/tex]. Therefore, [tex]\( a = \sqrt{36} = 6 \)[/tex] and [tex]\( b = \sqrt{64} = 8 \)[/tex].
4. Find the slopes of the asymptotes:
The asymptotes of the hyperbola [tex]\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = -1\)[/tex] are given by the equations:
[tex]\[ y = \pm \frac{b}{a} x \][/tex]
Calculate [tex]\(\frac{b}{a}\)[/tex]:
[tex]\[ \frac{b}{a} = \frac{8}{6} = \frac{4}{3} \][/tex]
5. Write the equations of the asymptotes:
Therefore, the equations of the asymptotes are:
[tex]\[ y = \frac{4}{3} x \quad \text{and} \quad y = -\frac{4}{3} x \][/tex]
So, the correct answer is:
[tex]\[ \boxed{y = \frac{4}{3} x \text{ and } y = -\frac{4}{3} x} \][/tex]