You are correct, I got the same foci
The one you looked at in your text probably had the y-axis as its major axis, that is, the focal points were on the y axis.
Was the denominator of the x^2 smaller than the denominator of the y^2 term?
If so, that was the case.
I am not too sure about this problem. Can you please help me?
Graph the ellipse and locate the foci.
9x^2=144-16y^2
I got (- squareroot 7, 0) and (squareroot 7,0), but I was looking at one problem in one book similar to it, and it had the answer as (0, - squareroot 7) and (0, squareroot 7).
4 answers
The one in book, is 7x^2=35-5y^2
Rewrite as
x^2/(144/9) + y^2/(144/16) = 1
(x/4)^2 + (y/3)^2 = 1
The semimajor and semiminor axis distances are 4 and 3, respectively. The major axis lies along the x axis. The distance from the center to the foci is sqrt (4^2 - 3^2) = sqrt 7
x^2/(144/9) + y^2/(144/16) = 1
(x/4)^2 + (y/3)^2 = 1
The semimajor and semiminor axis distances are 4 and 3, respectively. The major axis lies along the x axis. The distance from the center to the foci is sqrt (4^2 - 3^2) = sqrt 7
7x^2=35-5y^2 rewritten in standard form is
x^2/5 + y^2/7 = 1
sure enough the y term denominator is 7 which is larger than the 5 under the x^2.
Which means the y axis is the major axis like I said before and the focal points are on the y-axis
x^2/5 + y^2/7 = 1
sure enough the y term denominator is 7 which is larger than the 5 under the x^2.
Which means the y axis is the major axis like I said before and the focal points are on the y-axis
1 answer