What is ((x-3)^2}{49}+\frac{(y-5)^2)/4 =1 ?

The solution to ((x-3)^2}{49}+\frac{(y-5)^2)/4 =1 is Ellipse with (h,k)=(3,5),a=7,b=2

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((x-3)^2}{49}+\frac{(y-5)^2)/4 =1

\frac{( x - 3 ) ^{2}}{49} + \frac{( y - 5 ) ^{2}}{4} = 1

(h, k) = (3, 5), a = 7, b = 2

Solution steps

\frac{( x - 3 ) ^{2}}{49} + \frac{( y - 5 ) ^{2}}{4} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{49} + \frac{( y - 5 ) ^{2}}{4} = 1

in the form of the standard ellipse equation

\frac{( x - 3 ) ^{2}}{7 ^{2}} + \frac{( y - 5 ) ^{2}}{2 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (3, 5), a = 7, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (3, 5), a = 7, b = 2

8 x^{2} + 2 y^{2} - 16 x - 16 y + 8 = 0

12 x^{2} + 3 + 7 y^{2} = 5

5 x^{2} + 2 y^{2} + 12 y + 8 = 0

4 x^{2} + 5 y^{2} - 32 x + 30 y - 11 = 0

\frac{( x - 4 ) ^{2}}{100} + \frac{( y + 5 ) ^{2}}{49} = 1

What is ((x-3)^2}{49}+\frac{(y-5)^2)/4 =1 ?
The solution to ((x-3)^2}{49}+\frac{(y-5)^2)/4 =1 is Ellipse with (h,k)=(3,5),a=7,b=2