What is ((x-2)^2)/9+((y-3)^2)/(49)=1 ?

The solution to ((x-2)^2)/9+((y-3)^2)/(49)=1 is Ellipse with (h,k)=(2,3),a=3,b=7

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((x-2)^2)/9+((y-3)^2)/(49)=1

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{49} = 1

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

Solution steps

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{49} = 1

Rewrite

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{49} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

\frac{( x - 1 ) ^{2}}{8} + \frac{( y + \frac{3}{2} ) ^{2}}{4} = 1

11 x^{2} + 9 y^{2} = 1

x^{2} + y^{2} + 36 y = 0

x^{2} + 9 y^{2} - 2 x + 72 y + 136 = 0

36 x^{2} + 25 y^{2} - 50 y - 875 = 0

What is ((x-2)^2)/9+((y-3)^2)/(49)=1 ?
The solution to ((x-2)^2)/9+((y-3)^2)/(49)=1 is Ellipse with (h,k)=(2,3),a=3,b=7