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

The solution to (x^2}{25}+\frac{(y-2)^2)/4 =1 is Ellipse with (h,k)=(0,2),a=5,b=2

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

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

(h, k) = (0, 2), a = 5, b = 2

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{5 ^{2}} + \frac{( y - 2 ) ^{2}}{2 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 2), a = 5, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 2), a = 5, b = 2

\frac{( x + 7 ) ^{2}}{16} + \frac{( y + 5 ) ^{2}}{4} = 1

y^{2} = 9 - 4 x^{2}

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

\frac{x ^{2}}{3 ^{2}} + \frac{y ^{2}}{7 ^{2}} = 1

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

What is (x^2}{25}+\frac{(y-2)^2)/4 =1 ?
The solution to (x^2}{25}+\frac{(y-2)^2)/4 =1 is Ellipse with (h,k)=(0,2),a=5,b=2