What is ((x-6)^2)/(100)+((y+3)^2)/(16)=1 ?

The solution to ((x-6)^2)/(100)+((y+3)^2)/(16)=1 is Ellipse with (h,k)=(6,-3),a=10,b=4

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((x-6)^2)/(100)+((y+3)^2)/(16)=1

\frac{( x - 6 ) ^{2}}{100} + \frac{( y + 3 ) ^{2}}{16} = 1

(h, k) = (6, - 3), a = 10, b = 4

Solution steps

\frac{( x - 6 ) ^{2}}{100} + \frac{( y + 3 ) ^{2}}{16} = 1

Rewrite

\frac{( x - 6 ) ^{2}}{100} + \frac{( y + 3 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

\frac{( x - 6 ) ^{2}}{1 0 ^{2}} + \frac{( y - ( - 3 ) ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (6, - 3), a = 10, b = 4

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (6, - 3), a = 10, b = 4

2 x^{2} + 4 y^{2} + 9 x + 9 y - 17 = 0

(\frac{( x - 1 ) ^{2}}{4}) + (\frac{( y - 2 ) ^{2}}{9}) = 1

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

3 x^{2} + 2 y^{2} = 35

x^{2} + 5 y^{2} - 4 x - 16 = 0

What is ((x-6)^2)/(100)+((y+3)^2)/(16)=1 ?
The solution to ((x-6)^2)/(100)+((y+3)^2)/(16)=1 is Ellipse with (h,k)=(6,-3),a=10,b=4