What is ((x-2)^2)/9+((y-3)^2)/(16)=1 ?

The solution to ((x-2)^2)/9+((y-3)^2)/(16)=1 is Ellipse with (h,k)=(2,3),a=3,b=4

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((x-2)^2)/9+((y-3)^2)/(16)=1

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{16} = 1

(h, k) = (2, 3), a = 3, b = 4

Solution steps

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{16} = 1

Rewrite

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (2, 3), a = 3, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (2, 3), b = 4, a = 3

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

9 x^{2} + 36 y^{2} - 54 x + 72 y - 207 = 0

4 x^{2} + y^{2} - 48 x + 4 y + 48 = 0

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

4 x^{2} + y^{2} - 36 x + 120 y + 80 = 0

What is ((x-2)^2)/9+((y-3)^2)/(16)=1 ?
The solution to ((x-2)^2)/9+((y-3)^2)/(16)=1 is Ellipse with (h,k)=(2,3),a=3,b=4