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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

25 x^{2} + 16 y^{2} + 100 x - 96 y - 154 = 0

\frac{( x + 1 ) ^{2}}{25} + \frac{( y + 5 ) ^{2}}{16} = 1

49 x^{2} + 25 y^{2} - 1225 = 0

15 x^{2} + 12 y^{2} + 30 x - 24 y = 150

2 x^{2} + 5 y^{2} - 4 x + y - 6 = 0

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