What is ((x-3)^2)/4+((y+1)^2)/(25)=1 ?

The solution to ((x-3)^2)/4+((y+1)^2)/(25)=1 is Ellipse with (h,k)=(3,-1),a=2,b=5

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((x-3)^2)/4+((y+1)^2)/(25)=1

\frac{( x - 3 ) ^{2}}{4} + \frac{( y + 1 ) ^{2}}{25} = 1

(h, k) = (3, - 1), a = 2, b = 5

Solution steps

\frac{( x - 3 ) ^{2}}{4} + \frac{( y + 1 ) ^{2}}{25} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{4} + \frac{( y + 1 ) ^{2}}{25} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (3, - 1), a = 2, b = 5

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, - 1), b = 5, a = 2

4 x^{2} - 16 x + 9 y^{2} + 54 y + 61 = 0

9 x^{2} + 16 y^{2} - 18 x + 64 y = 71

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

\frac{x ^{2}}{\frac{25}{4}} + \frac{( y + \frac{5}{2} ) ^{2}}{\frac{9}{4}} = 1

\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = y + 6

What is ((x-3)^2)/4+((y+1)^2)/(25)=1 ?
The solution to ((x-3)^2)/4+((y+1)^2)/(25)=1 is Ellipse with (h,k)=(3,-1),a=2,b=5