What is ((x-2)^2}{16}+\frac{(y-1)^2)/7 =1 ?

The solution to ((x-2)^2}{16}+\frac{(y-1)^2)/7 =1 is Ellipse with (h,k)=(2,1),a=4,b=sqrt(7)

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

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

(h, k) = (2, 1), a = 4, b = 7

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

\frac{( x - 2 ) ^{2}}{4 ^{2}} + \frac{( y - 1 ) ^{2}}{( 7 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (2, 1), a = 4, b = 7

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (2, 1), a = 4, b = 7

6 x^{2} - 6 x + 3 y^{2} + 3 y - 102 = 0

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

x^{2} 8 + \frac{y ^{2}}{7 ^{2}} = 1

\frac{x ^{2}}{24} + \frac{y ^{2}}{49} = 1

\frac{x ^{2}}{64} + \frac{y ^{2}}{121} = 1

What is ((x-2)^2}{16}+\frac{(y-1)^2)/7 =1 ?
The solution to ((x-2)^2}{16}+\frac{(y-1)^2)/7 =1 is Ellipse with (h,k)=(2,1),a=4,b=sqrt(7)