What is ((x+2)^2}{20}+\frac{(y-3)^2)/4 =1 ?

The solution to ((x+2)^2}{20}+\frac{(y-3)^2)/4 =1 is Ellipse with (h,k)=(-2,3),a=2sqrt(5),b=2

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((x+2)^2}{20}+\frac{(y-3)^2)/4 =1

\frac{( x + 2 ) ^{2}}{20} + \frac{( y - 3 ) ^{2}}{4} = 1

(h, k) = (- 2, 3), a = 25, b = 2

Solution steps

\frac{( x + 2 ) ^{2}}{20} + \frac{( y - 3 ) ^{2}}{4} = 1

Rewrite

\frac{( x + 2 ) ^{2}}{20} + \frac{( y - 3 ) ^{2}}{4} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 2, 3), a = 25, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 2, 3), a = 25, b = 2

- 4 x^{2} - 9 y^{2} + 8 x - 18 y + 23 = 0

\frac{16 x ^{2}}{144} + \frac{9 y ^{2}}{144} = 1

x^{2} + 3 y^{2} - 13 - y^{2} - 4 x = 0

2 x^{2} - 200 x + y^{2} + 4 y + 379 = 0

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

What is ((x+2)^2}{20}+\frac{(y-3)^2)/4 =1 ?
The solution to ((x+2)^2}{20}+\frac{(y-3)^2)/4 =1 is Ellipse with (h,k)=(-2,3),a=2sqrt(5),b=2