What is ((x-1)^2)/(25)+((y+3)^2)/(100)=1 ?

The solution to ((x-1)^2)/(25)+((y+3)^2)/(100)=1 is Ellipse with (h,k)=(1,-3),a=5,b=10

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((x-1)^2)/(25)+((y+3)^2)/(100)=1

\frac{( x - 1 ) ^{2}}{25} + \frac{( y + 3 ) ^{2}}{100} = 1

(h, k) = (1, - 3), a = 5, b = 10

Solution steps

\frac{( x - 1 ) ^{2}}{25} + \frac{( y + 3 ) ^{2}}{100} = 1

Rewrite

\frac{( x - 1 ) ^{2}}{25} + \frac{( y + 3 ) ^{2}}{100} = 1

in the form of the standard ellipse equation

\frac{( x - 1 ) ^{2}}{5 ^{2}} + \frac{( y - ( - 3 ) ) ^{2}}{1 0 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (1, - 3), a = 5, b = 10

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (1, - 3), b = 10, a = 5

\frac{x ^{2}}{1} + \frac{y ^{2}}{4} = 1

- \frac{x ^{2}}{4} - \frac{z ^{2}}{4} = 1

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} \leq 1

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

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

What is ((x-1)^2)/(25)+((y+3)^2)/(100)=1 ?
The solution to ((x-1)^2)/(25)+((y+3)^2)/(100)=1 is Ellipse with (h,k)=(1,-3),a=5,b=10