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

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

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

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

(h, k) = (2, 3), a = 4, b = 5

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (2, 3), a = 4, b = 5

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (2, 3), b = 5, a = 4

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

\frac{x ^{2}}{49} + \frac{y ^{2}}{81} = 1

3 x^{2} + 5 y^{2} - 6 x + 2 y - 9 = 0

9 x^{2} + 25 y^{2} + 18 x - 100 y - 116 = 0

\frac{x ^{2}}{24} + \frac{y ^{2}}{9} = 1

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