What is ((x-1)^2)/4+((y-5)^2)/(25)=1 ?

The solution to ((x-1)^2)/4+((y-5)^2)/(25)=1 is Ellipse with (h,k)=(1,5),a=2,b=5

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((x-1)^2)/4+((y-5)^2)/(25)=1

\frac{( x - 1 ) ^{2}}{4} + \frac{( y - 5 ) ^{2}}{25} = 1

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

Solution steps

\frac{( x - 1 ) ^{2}}{4} + \frac{( y - 5 ) ^{2}}{25} = 1

Rewrite

\frac{( x - 1 ) ^{2}}{4} + \frac{( y - 5 ) ^{2}}{25} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

\frac{x ^{2}}{36} + \frac{y ^{2}}{5.9 1 ^{2}} = 1

16 x^{2} + 25 y^{2} - 128 x + 100 y = 0

\frac{( ( x + 0 ) ^{2} )}{40} + \frac{( ( y + 0 ) ^{2} )}{32} = 1

\frac{( x - 5 ) ^{2}}{16} + \frac{( y + 15 ) ^{2}}{25} = 1

\frac{x}{x ^{2} + y ^{2}} = \frac{1}{2}

What is ((x-1)^2)/4+((y-5)^2)/(25)=1 ?
The solution to ((x-1)^2)/4+((y-5)^2)/(25)=1 is Ellipse with (h,k)=(1,5),a=2,b=5