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how do i wright this equation in Y= form? 25(x+3)^2+36(y+4)^2=900

0 votes

This conic is an ellipse im pretty sure.

asked Nov 30, 2013 in ALGEBRA 2 by harvy0496 Apprentice
reshown Nov 30, 2013 by goushi

2 Answers

0 votes

Given equation is

25(x+3)^2+36(y+4)^2 = 900

divide to each sides by 900.

(25(x+3)^2)/900+(36(y+4)^2)/900 = 900/900

(x+3)^2/36+(y+4)^2/25 = 1

(x-(-3))^2/6^2+(y-(-4)^2/5^2 = 1

Compare it to standerd form of ellipse horizontal(x-h)^2/a^2+(y-k)^2/b^2 = 1, a>b

Center (h,k) = (-3,-4) and a = 6, b = 5

y = k+√(b^2(1-(x-h)^2/a^2) will give upper half of the ellipse.

y = k-√(b^2(1-(x-h)^2/a^2) will give lower half of the ellipse.

y = -4+√25(1-(x+3)^2/36)

y = -4+√25(1-(x-3)/6)^2

y = -4+√25√(1-(x-3)/6)^2

y = -4+5(1-(x-3)/6)

y = -4+5-5/6(x-3)

y = 1-(5/6)x+15/6

y = (5/6)x+1+5/2

y = (5/6)x+7/2 ----------> (1)

y = k-√(b^2(1-(x-h)^2/a^2)

y = -4-√25(1-(x+3)^2/36)

y = -4-√25(1-(x-3)/6)^2

y = -4-√25√(1-(x-3)/6)^2

y = -4-5(1-(x-3)/6)

y = -4-5+5/6(x-3)

y = -9+(5/6)x-5/2

y = (5/6)x-23/2 --------------> (2)

Equtions (1) , (2) are Y form to the given ellipse equation.

answered Nov 30, 2013 by william Mentor

Solution of image is image.

0 votes

The equation image

image

image

image

Compare it to standard form of ellipse image

a 2 > b 2

If the larger denominator is under the "x" term, then the ellipse is horizontal.

center (h, k ) = (-3, -4)

a  = length of semi-major axis = 6

= length of semi-minor axis = 5.

image for y  gives two equations.

 

image will give upper half of the ellipse.

image will give lower half of ellipse.

 

image

image

image

image

image

Solution image.

answered Jun 5, 2014 by david Expert

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