What is 9(x+2)^2+4(y-1)^2-36=0 ?

The solution to 9(x+2)^2+4(y-1)^2-36=0 is Ellipse with (h,k)=(-2,1),a=2,b=3

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9(x+2)^2+4(y-1)^2-36=0

9 (x + 2)^{2} + 4 (y - 1)^{2} - 36 = 0

(h, k) = (- 2, 1), a = 2, b = 3

Solution steps

9 (x + 2)^{2} + 4 (y - 1)^{2} - 36 = 0

Rewrite

9 (x + 2)^{2} + 4 (y - 1)^{2} - 36 = 0

in the form of the standard ellipse equation

\frac{( x - ( - 2 ) ) ^{2}}{2 ^{2}} + \frac{( y - 1 ) ^{2}}{3 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 2, 1), a = 2, b = 3

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 2, 1), b = 3, a = 2

9 x^{2} + y^{2} - 54 x - 8 y + 88 = 0

2 x^{2} + 4 y^{2} = 12

- 4 y^{2} - z^{2} + 1 = 0

36 x^{2} + 25 y^{2} = 3600

2 (x + 4)^{2} + (y + 2)^{2} = 8

What is 9(x+2)^2+4(y-1)^2-36=0 ?
The solution to 9(x+2)^2+4(y-1)^2-36=0 is Ellipse with (h,k)=(-2,1),a=2,b=3