What is ((x-2)^2}{25}+\frac{(y+3)^2)/4 =1 ?

The solution to ((x-2)^2}{25}+\frac{(y+3)^2)/4 =1 is Ellipse with (h,k)=(2,-3),a=5,b=2

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((x-2)^2}{25}+\frac{(y+3)^2)/4 =1

\frac{( x - 2 ) ^{2}}{25} + \frac{( y + 3 ) ^{2}}{4} = 1

(h, k) = (2, - 3), a = 5, b = 2

Solution steps

\frac{( x - 2 ) ^{2}}{25} + \frac{( y + 3 ) ^{2}}{4} = 1

Rewrite

\frac{( x - 2 ) ^{2}}{25} + \frac{( y + 3 ) ^{2}}{4} = 1

in the form of the standard ellipse equation

\frac{( x - 2 ) ^{2}}{5 ^{2}} + \frac{( y - ( - 3 ) ) ^{2}}{2 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (2, - 3), a = 5, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (2, - 3), a = 5, b = 2

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

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

5 x^{2} + 2 y^{2} = 98

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

2 x^{2} + 3 y^{2} - 12 x + 12 y + 12 = 0

What is ((x-2)^2}{25}+\frac{(y+3)^2)/4 =1 ?
The solution to ((x-2)^2}{25}+\frac{(y+3)^2)/4 =1 is Ellipse with (h,k)=(2,-3),a=5,b=2