1.Write the equation of the conic section with the given properties:
A hyperbola with vertices(0,6)(0,-6)and asymptotes y=3/4x and y=-3/4x.
2.Write the equation of the conic section with the given properties:
An ellipse with vertices(0,-5)(0,5)and a minor axis of length 8.
Im having major trouble i got x^2=4r2 for 1 and x^2=49 for 2
#1. vertices indicate you have a vertical axis of symmetry. So, since the foci are evenly spaced from (0,0),
y^2/a^2 - x^2/b^2 = 1
The vertices are at y = ±6, so
y^2/36 - x^2/b^2 = 1
The asymptotes are y = ±(b/a)x, so b/a = 3/4, or b=8
see the info at
http://www.wolframalpha.com/input/?i=hyperbola+y^2%2F36+-+x^2%2F64+%3D+1
The ellipse is similar. The semi-major axis is a=5, the semi-minor axis is b=4. Since the vertices are in the y-axis, equidistant from (0,0),
x^2/16 + y^2/25 = 1
See
http://www.wolframalpha.com/input/?i=ellipse+x^2%2F16+%2B+y^2%2F25+%3D+1
I have no idea how you got your results. They are in no way the equations of conic sections. You have some serious reviewing to do.
To find the equations of the conic sections with the given properties, follow these steps:
1. Equation of a hyperbola:
A hyperbola with vertices (0,6) and (0,-6) implies the center of the hyperbola is at the origin (0,0). The distance between the center and the vertices is called the transverse axis, which in this case is 6 units.
The equation of the asymptotes provides information about the slope of the branches of the hyperbola. The slopes of the asymptotes are given as 3/4 and -3/4.
The standard equation of a hyperbola with a vertical transverse axis and the center at the origin is given by:
(y - k)^2 / a^2 - (x - h)^2 / b^2 = 1
where (h, k) represents the center, a is the distance from the center to the vertices, and b is the distance from the center to the foci.
From the given information:
Center = (0,0)
a = 6
Since the transverse axis is vertical, b will represent the distance between the center and the endpoints of the conjugate axis, and its value will be larger than a.
The slope of the asymptotes, 3/4 and -3/4, gives us the ratio between a and b for the hyperbola.
The ratio a / b = 3/4
Now, we can calculate the value of b:
b = a / (3/4) = 4a / 3 = 4 * 6 / 3 = 8
Therefore, the equation of the hyperbola is:
(y - 0)^2 / 6^2 - (x - 0)^2 / 8^2 = 1
Simplifying it:
y^2 / 36 - x^2 / 64 = 1
Hence, the equation of the hyperbola is y^2 / 36 - x^2 / 64 = 1.
2. Equation of an ellipse:
An ellipse with vertices (0,-5) and (0,5) suggests that the center of the ellipse is at the origin (0,0). The distance between the center and the vertices is called the major axis, which in this case is 5 units.
The minor axis is given a length of 8 units.
The standard equation of an ellipse with a vertical major axis and the center at the origin is given by:
(x - h)^2 / b^2 + (y - k)^2 / a^2 = 1
where (h, k) represents the center, a is the distance from the center to the vertices, and b is the distance from the center to the endpoints of the minor axis.
From the given information:
Center = (0,0)
a = 5
The length of the minor axis = 2b = 8 (this means b = 4)
Therefore, the equation of the ellipse is:
(x - 0)^2 / 4^2 + (y - 0)^2 / 5^2 = 1
Simplifying it:
x^2 / 16 + y^2 / 25 = 1
Hence, the equation of the ellipse is x^2 / 16 + y^2 / 25 = 1.
Note: Your answer for the hyperbola equation in question 1 is correct, but your answer for the ellipse equation in question 2 is incorrect.
1. To write the equation of a hyperbola with the given properties, we can start by using the standard form of a hyperbola equation:
(x-h)^2/a^2 - (y-k)^2/b^2 = 1
where (h, k) represents the center of the hyperbola, and a and b are the lengths of the semi-major and semi-minor axes, respectively.
Given that the hyperbola has vertices at (0, 6) and (0, -6), we can determine that the center is at (h, k) = (0, 0). The distance between the center and each vertex is the length of the semi-major axis, so in this case, a = 6.
The asymptotes of the hyperbola are given by the equations y = (3/4)x and y = (-3/4)x. The slope of each asymptote is b/a, where b is the length of the semi-minor axis. In this case, b/a = 3/4, so b = (3/4)a = (3/4)(6) = 4.5.
Plugging these values into the standard form equation, we get:
(x-0)^2/(6^2) - (y-0)^2/(4.5^2) = 1
Simplifying, we have:
x^2/36 - y^2/20.25 = 1
So, the equation of the hyperbola is:
x^2/36 - y^2/20.25 = 1
2. To write the equation of an ellipse with the given properties, we can use the standard form of an ellipse equation:
(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.
Given that the ellipse has vertices at (0, -5) and (0, 5), we can determine that the center is at (h, k) = (0, 0). The distance between the center and each vertex is the length of the semi-major axis, so in this case, a = 5.
The length of the minor axis is given as 8, so the length of the semi-minor axis is b = 8/2 = 4.
Plugging these values into the standard form equation, we get:
(x-0)^2/(5^2) + (y-0)^2/(4^2) = 1
Simplifying, we have:
x^2/25 + y^2/16 = 1
So, the equation of the ellipse is:
x^2/25 + y^2/16 = 1