Find the equations of the asymptotes of the hyperbola defined by the equation shown below. If necessary, round to the nearest tenth.
−64x^2+ 9y^2- 1024x + 144y-4096 = 0
Asymptotes: =
and y=
The equation can be written as
(y+8)^2/64 - (x+8)^2/9 = 1
which gives us
b = 8, a = 3
making the asymptotes y+8 = ±8/3 (x+8)
Simplifying, we get the equations of the asymptotes as:
y = (8/3)x - 32/3 and y = -(8/3)x - 56/3
Rounded to the nearest tenth:
y = 2.7x - 10.7 and y = -2.7x - 18.7
To find the equations of the asymptotes of the hyperbola defined by the equation:
−64x^2+ 9y^2- 1024x + 144y-4096 = 0
First, we need to rewrite the equation in standard form by completing the square for both x and y terms:
−64x^2+ 9y^2- 1024x + 144y - 4096 = 0
Rearranging the terms, we get:
(−64x^2 - 1024x) + (9y^2 + 144y) = 4096
Factoring out the common coefficients:
−64(x^2 + 16x) + 9(y^2 + 16y) = 4096
Completing the square for the x term:
−64(x^2 + 16x + 64) + 9(y^2 + 16y) = 4096 - 64 * 64
Simplifying:
−64(x + 8)^2 + 9(y^2 + 16y) = 2816
Completing the square for the y term:
−64(x + 8)^2 + 9(y^2 + 16y + 64) = 2816 + 9 * 64
Simplifying further:
−64(x + 8)^2 + 9(y + 8)^2 = 3376
Now, we can rewrite the equation in standard form:
−64(x + 8)^2/3376 + 9(y + 8)^2/3376 = 1
Comparing this equation to the standard form of a hyperbola:
(x - h)^2/a^2 - (y - k)^2/b^2 = 1
We can see that:
h = -8
a^2 = 3376
k = -8
b^2 = 3376/9
The equation of the asymptotes of a hyperbola in standard form is given by:
(y - k) = ± (b/a)(x - h)
Substituting the values we found:
(y + 8) = ± (√(3376/9)/√3376)(x + 8)
Simplifying:
(y + 8) ≈ ± 1.4(x + 8)
Therefore, the equations of the asymptotes are approximately:
y + 8 ≈ 1.4(x + 8) and y + 8 ≈ -1.4(x + 8)
To find the equations of the asymptotes of a hyperbola, we need to rewrite its equation in the standard form: (x-h)^2/a^2 - (y-k)^2/b^2 = 1 for a horizontal hyperbola or (y-k)^2/a^2 - (x-h)^2/b^2 = 1 for a vertical hyperbola.
Let's start by completing the square for x variables:
-64x^2 + 9y^2 - 1024x + 144y - 4096 = 0
Rearrange the terms:
-64x^2 - 1024x + 9y^2 + 144y = 4096
Group the x and y terms separately:
(-64x^2 - 1024x) + (9y^2 + 144y) = 4096
Complete the square for the x terms:
-64(x^2 + 16x) + 9y^2 + 144y = 4096
To complete the square for x, we need to add the square of (half of the coefficient of x) and subtract it accordingly. In this case, (half of 16) is 8, and (8^2) is 64. When we add and subtract 64, the equation becomes:
-64(x^2 + 16x + 64 - 64) + 9y^2 + 144y = 4096
Simplify:
-64(x + 8)^2 + 9y^2 + 144y = 4096 + 64(64)
-64(x + 8)^2 + 9y^2 + 144y = 4096 + 4096
-64(x + 8)^2 + 9(y^2 + 16y) = 8192
Now, we need to complete the square for the y terms. Similarly, we will add and subtract the square of (half of the coefficient of y). In this case, (half of 16) is 8, and (8^2) is 64. When we add and subtract 64, the equation becomes:
-64(x + 8)^2 + 9(y^2 + 16y + 64 - 64) = 8192
Simplify:
-64(x + 8)^2 + 9(y^2 + 16y + 64) = 8192
-64(x + 8)^2 + 9(y + 8)^2 = 8192
Divide both sides by 8192 to normalize the equation:
-64(x + 8)^2/8192 + 9(y + 8)^2/8192 = 1
Simplify:
-(x + 8)^2/128 + (y + 8)^2/1024 = 1
Now, we have the equation of a vertical hyperbola:
(y + 8)^2/1024 - (x + 8)^2/128 = 1
The standard form of the equation tells us that:
h = -8 (horizontal shift of the center)
k = -8 (vertical shift of the center)
a^2 = 1024 (square of the distance from the center to the vertex along the y-axis)
b^2 = 128 (square of the distance from the center to the vertex along the x-axis)
To find the equations of the asymptotes, we can use the formula:
For a vertical hyperbola:
y = k ± (b/a)(x-h)
Substituting the values:
y = -8 ± (√(128/1024))(x + 8)
Simplifying:
y = -8 ± (0.4)(x + 8)
The equations of the asymptotes are:
y = -8 + 0.4(x + 8) and y = -8 - 0.4(x + 8)
Simplifying further, we have:
y = 0.4x + 5.2 and y = -0.4x - 11.2
Rounding to the nearest tenth, the equations of the asymptotes are:
y = 0.4x + 5.2 and y = -0.4x - 11.2.