Give the coordinates of the center, foci, vertices, and asymptotes of the hyperbola with the equation 4(x + 2) 2 + (y+4) 2 4 = 1
.Sketch the graph, and include these points and lines, along with the auxiliary rectangle.
Cannot sketch graphs on these posts.
You need to work on writing algebra online
4(x + 2)^2 + (y+4)^2/4 = 1
That's not the standard form, which would be
(x+2)^2 / (1/2)^2 + (y+4)^2/2^2 = 1
The other problem is, that this is the equation of an ellipse.
So, fix your equation and maybe we can get a useful answer
I will just remind you that
(x-h)^2/a^2 - (y-k)^2/b^2 = 1 has
c&^2 = a^2 + b^2
center at (h,k)
vertices at (h±a,k)
asymptotes y-k = ±b/a (x-h)
foci at (h±c,k)
To find the coordinates of the center, foci, vertices, and asymptotes of the given hyperbola, we can first rewrite the equation in a standard form.
The given equation of the hyperbola is 4(x + 2)^2 + (y + 4)^2/4 = 1.
1. Find the center:
The equation is in the form (x - h)^2/a^2 - (y - k)^2/b^2 = 1, where (h, k) represents the center.
Comparing this with the given equation, we have:
(x + 2)^2/1 - (y + 4)^2/4 = 1
So, the center of the hyperbola is at (-2, -4).
2. Find the vertices:
The vertices form the endpoints of the transverse axis, which is the axis passing through the center. The distance between the center and the vertices is given by a, where a represents the distance from the center to the vertex.
Since (x + 2) is squared and has a coefficient of 1, we know that a = 1. Therefore, the distance from the center to the vertex is 1.
So, the coordinates of the vertices are (-2 + 1, -4) and (-2 - 1, -4), which simplifies to (-1, -4) and (-3, -4) respectively.
3. Find the foci:
The foci are located along the transverse axis, and their distance from the center is given by c, where c^2 = a^2 + b^2.
In the given equation, a = 1. Substitute this value into the formula to calculate c:
c^2 = 1^2 + 2^2
c^2 = 1 + 4
c^2 = 5
c = √5
So, the distance from the center to the foci is √5 units.
The coordinates of the foci along the transverse axis are (-2 + √5, -4) and (-2 - √5, -4).
4. Find the asymptotes:
The slopes of the asymptotes are given by b/a, where b represents the distance from the center to the point on the conjugate axis.
In the given equation, b = 2. Therefore, the slope of the asymptotes is 2/1 = 2.
Using the point-slope form of a line, we can find the equations of the asymptotes.
For the hyperbola with its center at (-2, -4), the two equations of the asymptotes are:
y - (-4) = 2(x - (-2))
y + 4 = 2(x + 2)
y + 4 = 2x + 4
y = 2x
Now that we have the necessary coordinates, we can sketch the graph of the hyperbola with the center, foci, vertices, and asymptotes on it.
(Note: You can refer to an online graphing tool or software to draw an accurate graph of the hyperbola.)
To create an auxiliary rectangle, we consider the length of the transverse axis as the length of the rectangle, and the length of the conjugate axis as the width of the rectangle.
The length of the transverse axis is 2a, which is 2 units in this case. The length of the conjugate axis is 2b, which is 4 units in this case.
Therefore, the auxiliary rectangle can be drawn with its center at (-2, -4) and the lengths of 2 units (transverse axis) and 4 units (conjugate axis).