Please show all the work/break down
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=
To find the equations of the asymptotes of a hyperbola, we first need to write the equation in standard form, which is:
(x-h)^2/a^2 - (y-k)^2/b^2 = 1
where (h,k) is the center of the hyperbola and a and b are the distances from the center to the vertices along the x-axis and y-axis, respectively.
To get the given equation in standard form, we need to complete the square for both x and y terms:
-64(x^2 - 16x) + 9(y^2 + 16y) = 4096
-64(x^2 - 16x + 64) + 9(y^2 + 16y + 64/9) = 4096 - 64(64) + 9(64/9)
-64(x-8)^2 + 9(y+8/3)^2 = -384
Dividing both sides by -384, we get:
(x-8)^2/6 - (y+8/3)^2/48 = 1
Now we can see that the center of the hyperbola is (8,-8/3), a = sqrt(6) and b = sqrt(48) = 4sqrt(3).
The equation of the asymptotes for a hyperbola of the form (x-h)^2/a^2 - (y-k)^2/b^2 = 1 is given by:
y - k = (b/a) (x - h)
y - k = (-b/a) (x - h)
For our hyperbola, the asymptotes will pass through the center (8,-8/3) and have a slope of ±b/a.
Slope of the first asymptote:
m = 4sqrt(3)/sqrt(6) = 4sqrt(2)
Using point-slope form with (8,-8/3):
y + 8/3 = 4sqrt(2) (x - 8)
Rearranging, we get:
y = 4sqrt(2)x - 8sqrt(2) - 8/3
Slope of the second asymptote:
m = -4sqrt(3)/sqrt(6) = -4sqrt(2)
Using point-slope form with (8,-8/3):
y + 8/3 = -4sqrt(2) (x - 8)
Rearranging, we get:
y = -4sqrt(2)x + 8sqrt(2) - 8/3
Therefore, the equations of the asymptotes are:
y = 4sqrt(2)x - 8sqrt(2) - 8/3
y = -4sqrt(2)x + 8sqrt(2) - 8/3
A parabola can be drawn given a focus of (-2,7)
and a directrix of x=−10. What can be said about the parabola? The parabola has a vertex at....has a p-value of...and it
Since the directrix is a vertical line at x = -10, we know that the parabola opens horizontally to the right. The vertex will be halfway between the focus and the directrix, so its x-coordinate is the average of -2 and -10:
x-coordinate of vertex = (-2 + (-10))/2 = -6
The y-coordinate of the vertex is simply the y-coordinate of the focus:
y-coordinate of vertex = 7
So the vertex is at (-6, 7).
For a parabola opening horizontally, the distance from the vertex to the focus (and also to the directrix) is called the p-value. We can find this distance using the formula:
p = |(x-coordinate of focus) - (x-coordinate of vertex)|
p = |-2 - (-6)| = 4
So the parabola has a p-value of 4.
Putting all of this together, we can say that the parabola opens horizontally to the right, has a vertex at (-6, 7), and has a p-value of 4.
which way does it open up?
A parabola with a vertical axis of symmetry opens up if it has a minimum point and opens down if it has a maximum point. However, for a parabola with a horizontal axis of symmetry (which is the case in this problem), the parabola opens horizontally to the right if it has a minimum point and to the left if it has a maximum point.
Therefore, based on the information given in the problem, we cannot determine whether the parabola opens up or down since it is horizontally oriented.
Find the foci of the ellipse defined by the equation shown below. If necessary, round to the nearest tenth.
16x^2+ 9y^2+160x - 72 +400 = 0
Foci:
We first need to get the given equation into standard form, which is:
(x-h)^2/a^2 + (y-k)^2/b^2 = 1
where (h,k) is the center of the ellipse and a and b are the lengths of the semi-major and semi-minor axes, respectively.
To do this, we can complete the square for both the x and y terms:
16(x^2 + 10x) + 9(y^2 - 8) = -400
16(x^2 + 10x + 25) + 9(y^2 - 8) = -400 + 16(25)
16(x + 5)^2 + 9(y^2 - 8) = 0
Dividing both sides by -144, we get:
(x + 5)^2/9 + (y^2 - 8)/16 = 1
Now we can see that the center of the ellipse is (-5,0), a = 3 and b = 4.
The distance from the center to each focus is given by:
c = sqrt(a^2 - b^2)
c = sqrt(9 - 16) = sqrt(-7)
Since the radicand is negative, we know that the ellipse is imaginary and has no real foci.