What is ((x-2)/(4^2))^2+((y-4)/(5^2))^2=1 ?

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

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((x-2)/(4^2))^2+((y-4)/(5^2))^2=1

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

(h, k) = (2, 4), a = 16, b = 25

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

\frac{( x - 2 ) ^{2}}{1 6 ^{2}} + \frac{( y - 4 ) ^{2}}{2 5 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (2, 4), a = 16, b = 25

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (2, 4), b = 25, a = 16

9 x^{2} - 54 x + y^{2} - 2 y = - 81

\frac{( x + 2 ) ^{2}}{64} + \frac{( y - 1 ) ^{2}}{81} = 1

3 x^{2} + y^{2} = 52

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

8 x^{2} + 16 x + 4 y^{2} - 12 y = 15

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