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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

\frac{( x + 1 ) ^{2}}{16} + \frac{( y + 3 ) ^{2}}{9} = 1

x^{2} + (8 - 6) y^{2} = 1

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

\frac{100}{1 + x ^{2} + 2 y ^{2}} = 1

81 x^{2} + 49 y^{2} + 486 x - 490 y - 2015 = 0

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