What is ((x-2)^2)/(64)+((y-4)^2)/(16)=1 ?

The solution to ((x-2)^2)/(64)+((y-4)^2)/(16)=1 is Ellipse with (h,k)=(2,4),a=8,b=4

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((x-2)^2)/(64)+((y-4)^2)/(16)=1

\frac{( x - 2 ) ^{2}}{64} + \frac{( y - 4 ) ^{2}}{16} = 1

(h, k) = (2, 4), a = 8, b = 4

Solution steps

\frac{( x - 2 ) ^{2}}{64} + \frac{( y - 4 ) ^{2}}{16} = 1

Rewrite

\frac{( x - 2 ) ^{2}}{64} + \frac{( y - 4 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

\frac{( x - 2 ) ^{2}}{8 ^{2}} + \frac{( y - 4 ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (2, 4), a = 8, b = 4

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (2, 4), a = 8, b = 4

\frac{( x + 3 ) ^{2}}{4} + \frac{( y - 5 ) ^{2}}{25} = 1

5 x^{2} + 9 y^{2} = 100

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

\frac{( x + 6 ) ^{2}}{7} + \frac{( y - 3 ) ^{2}}{16} = 1

4 x^{2} + y^{2} + 64 x - y + 201 = 0

What is ((x-2)^2)/(64)+((y-4)^2)/(16)=1 ?
The solution to ((x-2)^2)/(64)+((y-4)^2)/(16)=1 is Ellipse with (h,k)=(2,4),a=8,b=4