What is (x^2}{49}+\frac{(y-5)^2)/4 =1 ?

The solution to (x^2}{49}+\frac{(y-5)^2)/4 =1 is Ellipse with (h,k)=(0,5),a=7,b=2

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(x^2}{49}+\frac{(y-5)^2)/4 =1

\frac{x ^{2}}{49} + \frac{( y - 5 ) ^{2}}{4} = 1

(h, k) = (0, 5), a = 7, b = 2

Solution steps

\frac{x ^{2}}{49} + \frac{( y - 5 ) ^{2}}{4} = 1

Rewrite

\frac{x ^{2}}{49} + \frac{( y - 5 ) ^{2}}{4} = 1

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{7 ^{2}} + \frac{( y - 5 ) ^{2}}{2 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 5), a = 7, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 5), a = 7, b = 2

25 x^{2} + 5 y^{2} - 20 y - 216 = 0

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

\frac{8 x ^{2} - 12 x + 8 y ^{2} + 20 y + 17}{8} - 2 = 0

\frac{x ^{2}}{20} + \frac{y ^{2}}{25} = 1

16 x^{2} + 25 y^{2} - 32 x - 100 y - 284 = 0

What is (x^2}{49}+\frac{(y-5)^2)/4 =1 ?
The solution to (x^2}{49}+\frac{(y-5)^2)/4 =1 is Ellipse with (h,k)=(0,5),a=7,b=2