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Question 716150: Given the equation (x+3)^2/36+(y-1)^2/18=1
find: the center. I found (-3,1)
vertices (-3.-5) and (-3,7)
the end points of minor axis I only came up with 6 radical 2
the foci I got 3 radical 2 and length of major axis is 12
Are these correct?
I can't find the length of focal cord or its endpoints.
Please help
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
You got the center in the right place. But the vertices are at the ends of the major axis. With the absence of an  term in the equation, the major axis lines up with the  axis when the denominator in the term is larger than the denominator in the  term. 36 is bigger than 18, so the major axis has to be horizontal. -3 - 6 is -9, so one of your vertices is at (-9,1) and -3 + 6 = 3, so the other vertex is at (3,10).
As for the endpoints of the minor axis, take the square root of the smaller denominator. Since 18 is 9 times 2, the square root of 18 is  . So your endpoints are ) and ) .
Since  , the focal distance,  , is the distance from the center to the focus, and the foci lie on the major axis, so the foci are at: ) and )
The semi-major axis is 6, so the major axis has to be 12.
The measure of the semi-latus rectum (i.e. half of the focal chord) is given by  or in your case,  . Hence the length of the focal chord is 6.
The latus rectum, or focal chord if you will, is perpendicular to the major axis at the focus. So for your ellipse, the left hand focal chord has end points (-9,4) and (-9,-2). I'll leave the endpoints of the other focal chord in your capable hands.
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
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