SOLUTION: Question: Find the equation of the ellipise whose centers is (-4,-5), has a vertex at (2,-5) and a minor axis of length 6. (A) (x+4)^2/9 + (y+5)^2/36 = 1 (B) (x+4)^2/36 + (y+5

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Question: Find the equation of the ellipise whose centers is (-4,-5), has a vertex at (2,-5) and a minor axis of length 6. (A) (x+4)^2/9 + (y+5)^2/36 = 1 (B) (x+4)^2/36 + (y+5      Log On


   



Question 31795: Question: Find the equation of the ellipise whose centers is (-4,-5), has a vertex at (2,-5) and a minor axis of length 6.
(A) (x+4)^2/9 + (y+5)^2/36 = 1
(B) (x+4)^2/36 + (y+5)^2/9 =1
(C) none of these
(D) x^2/36 + y^2/9 =1
Thank you!

Answer by mukhopadhyay(490) About Me  (Show Source):
You can put this solution on YOUR website!
Standard equation of ellipse is:
(x-h)^2/a^2 + (y-k)^2/b^2 = 1 where (h,k) is the center, a=1/2(length of one axis), and b=1/2(length of another axis);
We know that (h,k) = (-4,-5)
So, the equation boils down to
(x+4)^2/a^2 + (y+5)^2/b^2 = 1;
Vertex is at (2,-5) and center is at (-4,-5). This says that the major axis is parallel to x-axis (because y-value for both of them is the same) and 1/2 of the length of the major axis 6 [6 comes from 2-(-4); difference of x-value between vertex and the center]
So, a=6 and b=3 (6/2=3)
Thus, the equation of the ellipse is
(x+4)^2/6^2 + (y+5)^2/3^2 = 1
=>(x+4)^2/36 + (y+5)^2/9 = 1
Correct answer is B.