What is ((x+1)^2)/(25)+((y-3)^2)/(36)=1 ?

The solution to ((x+1)^2)/(25)+((y-3)^2)/(36)=1 is Ellipse with (h,k)=(-1,3),a=5,b=6

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((x+1)^2)/(25)+((y-3)^2)/(36)=1

\frac{( x + 1 ) ^{2}}{25} + \frac{( y - 3 ) ^{2}}{36} = 1

(h, k) = (- 1, 3), a = 5, b = 6

Solution steps

\frac{( x + 1 ) ^{2}}{25} + \frac{( y - 3 ) ^{2}}{36} = 1

Rewrite

\frac{( x + 1 ) ^{2}}{25} + \frac{( y - 3 ) ^{2}}{36} = 1

in the form of the standard ellipse equation

\frac{( x - ( - 1 ) ) ^{2}}{5 ^{2}} + \frac{( y - 3 ) ^{2}}{6 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 1, 3), a = 5, b = 6

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 1, 3), b = 6, a = 5

\frac{( x + 4 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{16} = 1

(x^{2} + 10 x) + (4 y^{2} + 9 y) = 7

8 x^{2} - 176 x + 9 y^{2} - 18 y + 905 = 0

9 x^{2} + 25 y^{2} = 400

36 - 4 x^{2} - 9 y^{2} = 0

What is ((x+1)^2)/(25)+((y-3)^2)/(36)=1 ?
The solution to ((x+1)^2)/(25)+((y-3)^2)/(36)=1 is Ellipse with (h,k)=(-1,3),a=5,b=6