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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 2 ) ^{2}}{49} = 1

16 x^{2} + y^{2} + 96 x - 10 y = - 153

x^{2} + y^{2} - 25 = \frac{x + 25}{7} - y

\frac{( x - 4 ) ^{2}}{9} + \frac{( y - 2 ) ^{2}}{25} = 1

(y - 6)^{2} = 23 - \frac{23 ( x - 3 ) ^{2}}{39}

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