What is (x^2}{16}+\frac{(y-2)^2)/9 =1 ?

The solution to (x^2}{16}+\frac{(y-2)^2)/9 =1 is Ellipse with (h,k)=(0,2),a=4,b=3

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(x^2}{16}+\frac{(y-2)^2)/9 =1

\frac{x ^{2}}{16} + \frac{( y - 2 ) ^{2}}{9} = 1

(h, k) = (0, 2), a = 4, b = 3

Solution steps

\frac{x ^{2}}{16} + \frac{( y - 2 ) ^{2}}{9} = 1

Rewrite

\frac{x ^{2}}{16} + \frac{( y - 2 ) ^{2}}{9} = 1

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{4 ^{2}} + \frac{( y - 2 ) ^{2}}{3 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 2), a = 4, b = 3

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 2), a = 4, b = 3

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

11 x^{2} + 13 y^{2} - 143 = 0

x^{2} + 3 y^{2} - 12 y + 9 = 0

4 x^{2} + 9 y^{2} - 40 x + 36 y + 100 = 0

\frac{( x - 8 ) ^{2}}{64} + \frac{( y - 5 ) ^{2}}{5 ^{2}} = 1

What is (x^2}{16}+\frac{(y-2)^2)/9 =1 ?
The solution to (x^2}{16}+\frac{(y-2)^2)/9 =1 is Ellipse with (h,k)=(0,2),a=4,b=3