(y+5)^2/169 + (x-5)^2/49 =1
What are the coordinates of the foci, the ends of the major and minor axes and the end of each latus rectum.
I already had the;
V1(18,-5);V2(-8,-5)
B1(5,2);B2(5,-12)
F1(15.95,-5);F2(-15.95,-5)
I cabt find out the latus rectum and the others. Please help me thnk you
Wow. First, you have x and y mixed up. Maybe you wrote it wrong, or maybe it was given to you that way to trip you up.
Anyway, using the given equation
(y+5)^2/169 + (x-5)^2/49 =1
let's write it in the usual format
(x-5)^2/49 + (y+5)^2/169 = 1
It is clear that the major axis is vertical, so we have
center: (5,-5)
a = 13
b = 7
so, c = √120 = 2√30
the foci are thus at (5,-5±√120)
vertices (ends of major axis): (5,-5±13)
covertices (ends of minor axis): (5±7,-5)
the latus rectum is the line segment through a focus (y = ±√120) and perpendicular to the major axis, and has length 2b^2/a
So, the ends of the LR are (5 ± 98/13,-5±√120)