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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

\frac{x ^{2}}{\frac{81}{4}} + \frac{y ^{2}}{\frac{25}{16}} = 1

9 x^{2} + 16 y^{2} + 72 x + 128 y + 256 = 0

9 x^{2} + 4 y^{2} - 18 x - 16 y - 11 = 0

x^{2} + 4 y^{2} - 25 = 0

4 x^{2} + y^{2} + 2 y - 35 = 0

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