What is (((x-4)^2)}{36}+\frac{((y-2)^2))/9 =1 ?

The solution to (((x-4)^2)}{36}+\frac{((y-2)^2))/9 =1 is Ellipse with (h,k)=(4,2),a=6,b=3

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(((x-4)^2)}{36}+\frac{((y-2)^2))/9 =1

\frac{( ( x - 4 ) ^{2} )}{36} + \frac{( ( y - 2 ) ^{2} )}{9} = 1

(h, k) = (4, 2), a = 6, b = 3

Solution steps

\frac{( x - 4 ) ^{2}}{36} + \frac{( y - 2 ) ^{2}}{9} = 1

Rewrite

\frac{( x - 4 ) ^{2}}{36} + \frac{( y - 2 ) ^{2}}{9} = 1

in the form of the standard ellipse equation

\frac{( x - 4 ) ^{2}}{6 ^{2}} + \frac{( y - 2 ) ^{2}}{3 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (4, 2), a = 6, b = 3

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (4, 2), a = 6, b = 3

9 x^{2} + 16 y^{2} - 90 x - 160 y + 481 = 0

16 x^{2} + 49 y^{2} = 25

2 x^{2} + y^{2} + 8 y + 14 = 0

18 x^{2} + 4 y^{2} + 4 y - 72 x + 37 = 0

x^{2} - 2 x + 2 y^{2} - 8 y + 7 = 0

What is (((x-4)^2)}{36}+\frac{((y-2)^2))/9 =1 ?
The solution to (((x-4)^2)}{36}+\frac{((y-2)^2))/9 =1 is Ellipse with (h,k)=(4,2),a=6,b=3