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

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

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

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

(h, k) = (- 2, 4), a = 2, b = 5

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 2, 4), a = 2, b = 5

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 2, 4), b = 5, a = 2

\frac{x ^{2}}{8 ^{2}} + \frac{y ^{2}}{5 ^{2}} = 1

x^{2} - 4 x + 4 y^{2} + 4 y + 1 = 0

25 x^{2} + 49 y^{2} = 49

x^{2} + x + 2 y^{2} + y - 1 = 0

\frac{( x - 4 ) ^{2}}{5} + \frac{( y - 6 ) ^{2}}{9} = 1

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