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Question 390926: how do you isolate the y in an ellipse equation like (x+6)^2/4+(y-5)^2/16=1?
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! given: (x+6)^2/4+(y-5)^2/16=1
standard form of an ellipse, (x-h)^2/b^2+(y-K)^2/a^2,a>b
let me first describe what the ellipse looks like just from inspection of the expression
this is an ellipse elongated (major axis) in the vertical direction because the y-term has the larger denominator (a).
coordinates of the center are (-6,5)
a^2=16
a=4
b^2=4
b=2
here is how to solve for y
multiply given expression by 16
4(x+6)^2+(y-5)^2=16
(y-5)^2=16-4(x+6)^2
take the sqrt of both sides
y-5 =sqrt(16-4(x+6)^2)
y=5ħsqrt(16-4(x+6)^2)
here is how to check to see if this is correct
let x=-6, the x coordinate of the center
y=5ħsqrt(16-4(-6+6)^2)=5ħsqrt(16)
y=5ħ4=9 and 1, which are the y coordinates of the vertices of the given ellipse.
another point we could check is when x=-4
y=5ħsqrt((16-4(-4+6)^2)=5ħsqrt(16-16)=5(the y coordinate of minor axis on the right side
ans: the formula to use in isolating y in given ellipse is: y=5ħsqrt(16-4(x+6)^2)
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