What is ((x-3)^2)/(16)+((y-3)^2)/(25)=1 ?

The solution to ((x-3)^2)/(16)+((y-3)^2)/(25)=1 is Ellipse with (h,k)=(3,3),a=4,b=5

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((x-3)^2)/(16)+((y-3)^2)/(25)=1

\frac{( x - 3 ) ^{2}}{16} + \frac{( y - 3 ) ^{2}}{25} = 1

(h, k) = (3, 3), a = 4, b = 5

Solution steps

\frac{( x - 3 ) ^{2}}{16} + \frac{( y - 3 ) ^{2}}{25} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{16} + \frac{( y - 3 ) ^{2}}{25} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (3, 3), a = 4, b = 5

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, 3), b = 5, a = 4

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

5 x^{2} + 8 y^{2} = 40

5 x^{2} + 3 y^{2} = 15

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

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

What is ((x-3)^2)/(16)+((y-3)^2)/(25)=1 ?
The solution to ((x-3)^2)/(16)+((y-3)^2)/(25)=1 is Ellipse with (h,k)=(3,3),a=4,b=5