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

Question

This conic is an ellipse im pretty sure.

Answer 1

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.

Comments

Solution of is .

Answer 2

The equation

Compare it to standard form of ellipse

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

b  = length of semi-minor axis = 5.

for y  gives two equations.

 

will give upper half of the ellipse.

will give lower half of ellipse.

 

Solution .