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

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

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

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

(h, k) = (4, 5), a = 2, b = 6

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (4, 5), a = 2, b = 6

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (4, 5), b = 6, a = 2

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

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

5 x^{2} + 13 y^{2} - 65 = 0

y^{2} = 1 - 16 x^{2}

4 x^{2} + y^{2} + 12 x + 6 y + 5 = 0

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