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
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 answer