locate the center, foci, vertices,ends of the latera recta and draw the conic. Compute the eccentricity and equation of the conic.
1) 5x^2 + 2y^2 = 100
2)y^2 - 2x - 8y + 10 = 0
1) To determine the center, foci, vertices, ends of the latera recta, and draw the conic for the equation 5x^2 + 2y^2 = 100, we need to put it in standard form.
First, divide both sides of the equation by 100:
x^2/20 + y^2/50 = 1
Comparing this to the standard form of an ellipse, we can see that:
a^2 = 50, b^2 = 20
The center of the ellipse is (h, k), where h and k are the coordinates of the center. In this case, h = 0 and k = 0, so the center is (0, 0).
The vertices are located at (h ± a, k), so the vertices of the ellipse are (0 ± √50, 0), which simplifies to (√50, 0) and (-√50, 0).
To find the foci, we use the formula c^2 = a^2 - b^2, where c is the distance between the center and the foci. In this case, c^2 = 50 - 20 = 30, so c = √30. The foci are located at (h ± c, k), so the foci of the ellipse are (0 ± √30, 0), which simplifies to (√30, 0) and (-√30, 0).
The ends of the latera recta are located at (h ± b^2/a, k), so the ends of the latera recta for the ellipse are (0 ± (20/√50), 0), which simplifies to (√20, 0) and (-√20, 0).
To compute the eccentricity, we use the formula e = c/a. In this case, e = √30/√50 = √(30/50) = √(3/5).
Now, let's draw the conic:
(Note: Drawing the conic visual can't be done through plain text. Please refer to a graphing tool to see the visual representation of the ellipse.)
The equation of the conic is 5x^2 + 2y^2 = 100.
2) To determine the center, foci, vertices, ends of the latera recta, and draw the conic for the equation y^2 - 2x - 8y + 10 = 0, we need to put it in standard form.
Rearranging the equation, we get:
y^2 - 8y = 2x - 10
Completing the square on the left side, we have:
(y^2 - 8y + 16) = 2x - 10 + 16
(y - 4)^2 = 2x + 6
Comparing this to the standard form of a parabola, we can see that:
4p = 2, p = 1/2
The vertex of the parabola is (h, k), where h and k are the coordinates of the vertex. In this case, h = -6 and k = 4, so the vertex is (-6, 4).
The focus is located at (h + p, k), so the focus of the parabola is (-6 + 1/2, 4), which simplifies to (-11/2, 4).
The equation of the directrix is y = k - p. In this case, the equation of the directrix is y = 4 - 1/2, which simplifies to y = 7/2.
To find the ends of the latera recta, we use the formula y = k ± p. In this case, the ends of the latera recta for the parabola are 4 ± 1/2, which simplifies to 3 1/2 and 4 1/2.
Now, let's draw the conic:
(Note: Drawing the conic visual can't be done through plain text. Please refer to a graphing tool to see the visual representation of the parabola.)
The equation of the conic is y^2 - 2x - 8y + 10 = 0.
Please note that the eccentricity cannot be computed for a parabola since it is defined only for ellipses and hyperbolas.
To find the center, foci, vertices, and ends of the latus rectum, as well as draw the conic, compute the eccentricity, and find the equation of the conic, we need to analyze each equation separately.
1) 5x^2 + 2y^2 = 100:
Let's start by rearranging the equation to put it in standard form:
Divide both sides of the equation by 100:
(x^2)/20 + (y^2)/50 = 1
Now, we can see that this equation represents an ellipse.
The equation is in the form (x^2)/(a^2) + (y^2)/(b^2) = 1, where a and b are the semi-major and semi-minor axes, respectively.
To find the center, we equate the denominators of x and y to 0:
a^2 = 20 => a = √20 = 2√5
b^2 = 50 => b = √50 = 5√2
The center of the ellipse is at the origin (0, 0).
The vertices are located at (±a, 0):
Vertex V1: (2√5, 0)
Vertex V2: (-2√5, 0)
To find the foci, we use the equation c = √(a^2 - b^2):
c = √(20 - 50) = √(-30), which is imaginary.
Since the foci do not exist, the eccentricity cannot be computed.
Finally, we can draw the ellipse based on the information we have gathered. We know the center, vertices, and the axes' lengths, so we can plot these points and sketch the ellipse accordingly.
2) y^2 - 2x - 8y + 10 = 0:
Let's rearrange the equation to identify the conic:
y^2 - 8y = 2x - 10
Completing the square for the y terms:
(y^2 - 8y + 16) = 2x - 10 + 16
(y - 4)^2 = 2x + 6
Now, we can see that this equation represents a parabola.
The equation is in the form (y - k)^2 = 4a(x - h), where (h, k) is the vertex.
Comparing this to our equation, we have:
(k, h) = (4, -3) => Vertex: (-3, 4)
The focus of a parabola is located at a distance |a| from the vertex along the axis of symmetry. In this case, the axis of symmetry is vertical, and the vertex is (h, k) = (-3, 4).
Since a is not given, we cannot compute the focus directly.
To find the ends of the latus rectum, we calculate 4a:
4a = 4a(x - h)
4a = 4
The latus rectum is a line segment perpendicular to the axis of symmetry, passing through the focus, and having its endpoints on the parabola. It has the length 4a.
Since a = 1, the length of the latus rectum is 4.
Finally, we sketch the parabola based on the information we have gathered. We plot the vertex and estimate the shape of the curve, keeping in mind that it opens to the right.