What is (((x-5)^2))/9+(((y-4)^2))/(36)=1 ?

The solution to (((x-5)^2))/9+(((y-4)^2))/(36)=1 is Ellipse with (h,k)=(5,4),a=3,b=6

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(((x-5)^2))/9+(((y-4)^2))/(36)=1

\frac{( ( x - 5 ) ^{2} )}{9} + \frac{( ( y - 4 ) ^{2} )}{36} = 1

(h, k) = (5, 4), a = 3, b = 6

Solution steps

\frac{( x - 5 ) ^{2}}{9} + \frac{( y - 4 ) ^{2}}{36} = 1

Rewrite

\frac{( x - 5 ) ^{2}}{9} + \frac{( y - 4 ) ^{2}}{36} = 1

in the form of the standard ellipse equation

\frac{( x - 5 ) ^{2}}{3 ^{2}} + \frac{( y - 4 ) ^{2}}{6 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (5, 4), a = 3, b = 6

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (5, 4), b = 6, a = 3

15 x^{2} + 10 y^{2} = 150

2 x^{2} + y^{2} - 4 x - 4 y = 0

25 x^{2} + y^{2} + 100 x + 4 y + 79 = 0

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

\frac{x ^{2}}{( 1.587 ) ^{2}} + \frac{y ^{2}}{1} = 1

What is (((x-5)^2))/9+(((y-4)^2))/(36)=1 ?
The solution to (((x-5)^2))/9+(((y-4)^2))/(36)=1 is Ellipse with (h,k)=(5,4),a=3,b=6