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

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

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

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

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

Solution steps

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

1 x^{2} + 4 y^{2} + 10 x + 8 y - 19 = 0

36 x^{2} + 9 y^{2} + 288 x - 90 y + 477 = 0

(\frac{x ^{2}}{72}) + (\frac{y ^{2}}{21}) = 1

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

\frac{( x - 4 ) ^{2}}{9} + \frac{( y - 7 ) ^{2}}{36} = 1

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