What is ((x+3)^2)/(36)+((y-4)^2)/(16)=1 ?

The solution to ((x+3)^2)/(36)+((y-4)^2)/(16)=1 is Ellipse with (h,k)=(-3,4),a=6,b=4

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((x+3)^2)/(36)+((y-4)^2)/(16)=1

\frac{( x + 3 ) ^{2}}{36} + \frac{( y - 4 ) ^{2}}{16} = 1

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

Solution steps

\frac{( x + 3 ) ^{2}}{36} + \frac{( y - 4 ) ^{2}}{16} = 1

Rewrite

\frac{( x + 3 ) ^{2}}{36} + \frac{( y - 4 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

\frac{( x - ( - 3 ) ) ^{2}}{6 ^{2}} + \frac{( y - 4 ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 3, 4), a = 6, b = 4

2 (x - 1)^{2} + 3 (y - 2)^{2} = 6

10 - \frac{x ^{2}}{2} - y^{2} = x^{2} + 4 y^{2}

x^{2} + 1 - 2 x + 4 y^{2} + 16 y > 0

(x + 5)^{2} + 9 (y + 3)^{2} = 36

\frac{( x + 5 ) ^{2}}{4} + \frac{( y + 5 ) ^{2}}{25} = 1

What is ((x+3)^2)/(36)+((y-4)^2)/(16)=1 ?
The solution to ((x+3)^2)/(36)+((y-4)^2)/(16)=1 is Ellipse with (h,k)=(-3,4),a=6,b=4