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

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

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

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

(h, k) = (2, 1), a = 2, b = 4

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (2, 1), a = 2, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (2, 1), b = 4, a = 2

\frac{( x - 6 ) ^{2}}{36} + \frac{y ^{2}}{100} = 1

\frac{2 y}{x ^{2} + y ^{2}} < 1

\frac{x ^{2}}{4 ^{2}} + \frac{y ^{2}}{2} = 1

\frac{x ^{2}}{6 ^{2}} + \frac{y ^{2}}{7 ^{2}} = 1

9 x^{2} + 25 y^{2} + 18 x - 50 y - 191 = 0

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