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

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

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

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

(h, k) = (1, 2), a = 4, b = 3

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (1, 2), a = 4, b = 3

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (1, 2), a = 4, b = 3

x^{2} + 5 y^{2} = 9 (2.1)

81 x^{2} + 16 y^{2} = 1296

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

\frac{( x - 4 ) ^{2}}{24} + \frac{( y - 2 ) ^{2}}{49} = 1

5 x^{2} + 2 y^{2} - 10 x + 18 y - 10 = 0

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