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

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

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

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

(h, k) = (5, 4), a = 1, b = 2

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (5, 4), a = 1, b = 2

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (5, 4), b = 2, a = 1

9 x^{2} + 36 y^{2} + 54 x - 288 y + 333 = 0

25 x^{2} + 9 y^{2} - 10 y - 216 = 0

(36 - 4 x^{2} - 9 y^{2}) = 0

\frac{( x - 5 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{16} = 1

9 x^{2} + 25 y^{2} - 900 = 0

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