Find the coordinates of the foci, the ends of major and minor axis, the ends of latus rectum then sketch the graph of y^2/169 + x^2/144 =1
(i need the anwers to that i will have an idea on how to solve the remaining .)
the major axis is vertical, so we have
center: (0,0)
a=13, b=12
c^2 = a^2-b^2, so c=5
vertices: (0,±13)
covertices: (±12,0)
foci: (0,±5)
eccentricity e = c/a = 5/13
length of latus rectum is 2b^2/a = 2*144/13, so the latera recti are the lines through the foci with ends at
(±144/13,5) and (±144/13,-5)
http://www.wolframalpha.com/input/?i=ellipse+y^2%2F169+%2B+x^2%2F144+%3D1
To find the coordinates of the foci, ends of the major and minor axis, and the ends of the latus rectum for the given equation of an ellipse, y^2/169 + x^2/144 = 1, we can start by identifying the key parameters of the ellipse equation.
The general equation of an ellipse centered at the origin is (x^2/a^2) + (y^2/b^2) = 1, where a is the length of the semi-major axis and b is the length of the semi-minor axis.
In the given equation, y^2/169 + x^2/144 = 1, we can see that a^2 = 169 and b^2 = 144. To find the length of the semi-major axis (a) and the semi-minor axis (b), we take the square root of these values:
a = √169 = 13
b = √144 = 12
The coordinates of the foci can be found using the formula c = √(a^2 - b^2), where c is the distance from the center to the foci. Plugging in the values, we get:
c = √(13^2 - 12^2) = √(169 - 144) = √25 = 5
Since the center of the ellipse is at the origin, the foci will be located on the x-axis. The coordinates of the foci will be (±c, 0), which in this case are (±5, 0). Therefore, the foci are located at (-5, 0) and (5, 0).
The ends of the major axis are given by the endpoints where the ellipse intersects the x-axis. Since the equation is y^2/169 + x^2/144 = 1, when y = 0, we can solve for x:
0^2/169 + x^2/144 = 1
x^2/144 = 1
x^2 = 144
x = ± √144
x = ±12
Therefore, the ends of the major axis are (-12, 0) and (12, 0).
The ends of the minor axis are given by the endpoints where the ellipse intersects the y-axis. Since the equation is y^2/169 + x^2/144 = 1, when x = 0, we can solve for y:
y^2/169 + 0 = 1
y^2 = 169
y = ± √169
y = ±13
Therefore, the ends of the minor axis are (0, -13) and (0, 13).
The latus rectum is defined as the line segment through the foci that is perpendicular to the major axis and passes through the ellipse curve. Its length is given by the formula 2b^2/a.
Plugging in our values, we get:
2(12^2)/13 = 2(144)/13 = 288/13
Therefore, the length of the latus rectum is 288/13.
To sketch the graph of the given equation, plot the center at the origin (0,0), then plot the foci at (-5,0) and (5,0). Draw the major axis passing through the foci and the center, with endpoints at (-12,0) and (12,0). Draw the minor axis passing through the center with endpoints at (0,-13) and (0,13). Finally, sketch the ellipse curve connecting the endpoints of the major and minor axes.