What is ((x-1)^2)/(25)+((y+3)^2)/(5100)=1 ?

The solution to ((x-1)^2)/(25)+((y+3)^2)/(5100)=1 is Ellipse with (h,k)=(1,-3),a=5,b=10sqrt(51)

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((x-1)^2)/(25)+((y+3)^2)/(5100)=1

\frac{( x - 1 ) ^{2}}{25} + \frac{( y + 3 ) ^{2}}{5100} = 1

(h, k) = (1, - 3), a = 5, b = 1051

Solution steps

\frac{( x - 1 ) ^{2}}{25} + \frac{( y + 3 ) ^{2}}{5100} = 1

Rewrite

\frac{( x - 1 ) ^{2}}{25} + \frac{( y + 3 ) ^{2}}{5100} = 1

in the form of the standard ellipse equation

\frac{( x - 1 ) ^{2}}{5 ^{2}} + \frac{( y - ( - 3 ) ) ^{2}}{( 10 51 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (1, - 3), a = 5, b = 1051

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (1, - 3), b = 1051, a = 5

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

36 x^{2} + 16 y^{2} - 576 = 0

\frac{x ^{2}}{9} + \frac{2 x}{3} + 16 y^{2} - 64 y + \frac{569}{9} = 0

\frac{( x - 2 ) ^{2}}{4} + \frac{( y + 1 ) ^{2}}{16} = 1

\frac{( ( x + 0 ) ^{2} )}{20} + \frac{( ( y + 0 ) ^{2} )}{16} = 1

What is ((x-1)^2)/(25)+((y+3)^2)/(5100)=1 ?
The solution to ((x-1)^2)/(25)+((y+3)^2)/(5100)=1 is Ellipse with (h,k)=(1,-3),a=5,b=10sqrt(51)