The transformation from 1 to 2 is a translation, as the shape is moved from one point to another with no rotation, reflection, or change in size.
The transformation from 2 to 3 is a reflection, as the shape is now upside down. It is still the same distance from the y-axis.
The answer is D. translation, then reflection.
The foci need to be equal distances from the center of the eclipse.
See the attached picture. I marked off all the points that were given as choice and from this you can see that that (-5,-4) and (-5,2) are both 3 units away from the center.
The answer is: (-5,-4) and (-5,2)
solution:
Consider the differential equation,
7ty + (1+t2)1/2y1 = 0
Rewrite the DE as,
(1+t2)1/2dy = - 7ty
dy/y = -7t/√1+t2 dt
in y = -7(1+t2) + c
y = ce-7(1+t2)
given,
y(0) = 1 => ce-7 = -1 => c = e7
ᴪ (t,y) = y -ce ᴪ(0,1)(1+t2)
