What is ((x-1)^2)/(25)+((y+2)^2)/(16)=1 ?

The solution to ((x-1)^2)/(25)+((y+2)^2)/(16)=1 is Ellipse with (h,k)=(1,-2),a=5,b=4

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((x-1)^2)/(25)+((y+2)^2)/(16)=1

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

x^{2} + y^{2} + 10 y = 0

12 x^{2} + 96 x + 5 y^{2} + 50 y + 257 = 0

\frac{( x - 3 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{3} = 1

\frac{( x - 2 ) ^{2}}{81} + \frac{( y + 5 ) ^{2}}{64} = 1

\frac{( x + 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{25} = 1

What is ((x-1)^2)/(25)+((y+2)^2)/(16)=1 ?
The solution to ((x-1)^2)/(25)+((y+2)^2)/(16)=1 is Ellipse with (h,k)=(1,-2),a=5,b=4