Ellipses question: x^2/25 + y^2/169=1?
1. state lengths of major & minor axes
2. state the x-inter and y-inter
3. find coordinates of the foci
4. find the points of intersection with line y=1-2x
5. graph
6. find the equation of the ellipse after it is translated accoring to[(x,y) --> (x-3,y+1)]?
could someone help me with this. I have 4 other questions that look similar and im not sure if i'm doing it correctly?
1) since b>a, the major axis will be the y-axis ,and the minor axis is the x-axis.
2) let x=0 to get y = ±13, so the y-intercepts are (0,13) and (0,-13)
Do the x-intercepts the same way, letting y = 0
3) In this type of ellipse, a^2 + c^2 = b^2
25 + c^2 = 169
c = ±12
the foci are (0,12) and (0,-12)
4) sub (1-2x) for the y in the equation to get
169x^2 + 25(1 - 4x + 4x^2) = 25(169)
I am sure you can solve this. Visualizing the graph, there should be 2 points of intersection, one in the 2nd and one in the fourth quadrants.
5. You should be able to do this knowing the vertices.
6. according to [(x,y) --> (x-3,y+1)]
your centre of (0,0) will move to (-3,1)
so the new equation is
(x+3)^2 / 25 + (y-1)^2 / 169 = 1
Certainly! I'd be happy to help you with your questions regarding the given ellipse equation. Let's go through each question step by step:
1. To determine the lengths of the major and minor axes, we can compare the coefficients of x^2 and y^2 in the equation of the ellipse. The general form of an ellipse equation is (x-h)^2/a^2 + (y-k)^2/b^2 = 1, where (h,k) represents the center of the ellipse.
Comparing this with your given equation, x^2/25 + y^2/169 = 1, we can see that a^2 = 25 and b^2 = 169. Therefore, the length of the major axis is 2a = 2*sqrt(25) = 10, and the length of the minor axis is 2b = 2*sqrt(169) = 26.
2. To find the x-intercepts, we can set y = 0 in the equation of the ellipse and solve for x. Substituting y = 0 into the equation x^2/25 + y^2/169 = 1 yields x^2/25 = 1. Solving this equation, we get x = ±5, so the x-intercepts are (-5, 0) and (5, 0).
Similarly, to find the y-intercepts, we can set x = 0 in the equation of the ellipse and solve for y. Substituting x = 0 into the equation x^2/25 + y^2/169 = 1 gives y^2/169 = 1. Solving for y, we obtain y = ±13, so the y-intercepts are (0, -13) and (0, 13).
3. The coordinates of the foci can be found using the relationship between the length of the major axis and the distance between the center and the foci. The formula is c^2 = a^2 - b^2, where c represents the distance from the center to each focus.
In our case, a^2 = 25 and b^2 = 169. Plugging these values into the formula, we get c^2 = 25 - 169 = -144. Since the result is negative, there are no real foci for this ellipse.
4. To find the points of intersection between the ellipse and the line y = 1 - 2x, we can substitute y in the equation of the ellipse. Substituting y = 1 - 2x into the equation x^2/25 + y^2/169 = 1, we get the quadratic equation x^2/25 + (1-2x)^2/169 = 1. By solving this equation, you'll obtain the x-coordinates of the points of intersection. Substitute these x-values back into the equation y = 1 - 2x to find the corresponding y-coordinates.
5. To graph the ellipse, you can plot the center (h, k) of the ellipse, which can be identified by the coefficients in the equation (x-h)^2/a^2 + (y-k)^2/b^2 = 1. In this case, the center is at (0, 0).
Then, you can plot the x-intercepts (-5, 0) and (5, 0), and the y-intercepts (0, -13) and (0, 13). Finally, sketch the ellipse passing through these points.
6. To translate the ellipse according to the transformation (x,y) --> (x-3,y+1), you simply need to substitute the new coordinates into the original ellipse equation. The new equation will have the same form: ((x-3)^2)/a^2 + ((y+1)^2)/b^2 = 1. Replace a^2 and b^2 with their respective values from the original equation.
I hope this helps! If you have any further questions or if anything is unclear, please let me know.