What is ((y-4}{16})^2+(\frac{x-6)/4)^2=1 ?

The solution to ((y-4}{16})^2+(\frac{x-6)/4)^2=1 is Ellipse with (h,k)=(6,4),a=4,b=16

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((y-4}{16})^2+(\frac{x-6)/4)^2=1

(\frac{y - 4}{16})^{2} + (\frac{x - 6}{4})^{2} = 1

(h, k) = (6, 4), a = 4, b = 16

Solution steps

(\frac{y - 4}{16})^{2} + (\frac{x - 6}{4})^{2} = 1

Rewrite

(\frac{y - 4}{16})^{2} + (\frac{x - 6}{4})^{2} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (6, 4), a = 4, b = 16

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (6, 4), b = 16, a = 4

8 x^{2} + y^{2} - 32 x - 16 y = - 16

6 x^{2} + 3 y^{2} + 3 y + 6 x = 15

\frac{x ^{2}}{16} + \frac{y ^{2}}{48} = 1

9^{2} x^{2} + 7^{2} y^{2} = 1

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{5} = 1

What is ((y-4}{16})^2+(\frac{x-6)/4)^2=1 ?
The solution to ((y-4}{16})^2+(\frac{x-6)/4)^2=1 is Ellipse with (h,k)=(6,4),a=4,b=16