What is ((x-12)^2)/(100)+((y+7)^2)/(36)=1 ?

The solution to ((x-12)^2)/(100)+((y+7)^2)/(36)=1 is Ellipse with (h,k)=(12,-7),a=10,b=6

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((x-12)^2)/(100)+((y+7)^2)/(36)=1

\frac{( x - 12 ) ^{2}}{100} + \frac{( y + 7 ) ^{2}}{36} = 1

(h, k) = (12, - 7), a = 10, b = 6

Solution steps

\frac{( x - 12 ) ^{2}}{100} + \frac{( y + 7 ) ^{2}}{36} = 1

Rewrite

\frac{( x - 12 ) ^{2}}{100} + \frac{( y + 7 ) ^{2}}{36} = 1

in the form of the standard ellipse equation

\frac{( x - 12 ) ^{2}}{1 0 ^{2}} + \frac{( y - ( - 7 ) ) ^{2}}{6 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (12, - 7), a = 10, b = 6

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (12, - 7), a = 10, b = 6

12 x^{2} + 10 y^{2} = 120

25 x^{2} + 9 y^{2} - 50 x + 54 y - 119 = 0

x, x^{2} + 7 k + 3 k (x)^{2} = 0

16 x^{2} + 25 y^{2} + 64 x + 150 y - 111 = 0

\frac{( x - 1 ) ^{2}}{81} + \frac{( y - 2 ) ^{2}}{25} = 1

What is ((x-12)^2)/(100)+((y+7)^2)/(36)=1 ?
The solution to ((x-12)^2)/(100)+((y+7)^2)/(36)=1 is Ellipse with (h,k)=(12,-7),a=10,b=6