What is ((x+2)^2)/((-13)^2)+((y-3)^2)/((-8)^2)=1 ?

The solution to ((x+2)^2)/((-13)^2)+((y-3)^2)/((-8)^2)=1 is Ellipse with (h,k)=(-2,3),a=13,b=8

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((x+2)^2)/((-13)^2)+((y-3)^2)/((-8)^2)=1

\frac{( x + 2 ) ^{2}}{( - 13 ) ^{2}} + \frac{( y - 3 ) ^{2}}{( - 8 ) ^{2}} = 1

(h, k) = (- 2, 3), a = 13, b = 8

Solution steps

\frac{( x + 2 ) ^{2}}{( - 13 ) ^{2}} + \frac{( y - 3 ) ^{2}}{( - 8 ) ^{2}} = 1

Rewrite

\frac{( x + 2 ) ^{2}}{( - 13 ) ^{2}} + \frac{( y - 3 ) ^{2}}{( - 8 ) ^{2}} = 1

in the form of the standard ellipse equation

\frac{( x - ( - 2 ) ) ^{2}}{1 3 ^{2}} + \frac{( y - 3 ) ^{2}}{8 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 2, 3), a = 13, b = 8

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 2, 3), a = 13, b = 8

\frac{x ^{2}}{4} + \frac{y ^{2}}{9} = 25

9 x^{2} + 16 y^{2} + 54 x + 64 y + 1 = 0

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

\frac{x ^{2}}{4} + \frac{( y + 3 ) ^{2}}{9} = 1

\frac{1}{2} (3 x^{2} + y^{2}) - 2 (x - y) = 0

What is ((x+2)^2)/((-13)^2)+((y-3)^2)/((-8)^2)=1 ?
The solution to ((x+2)^2)/((-13)^2)+((y-3)^2)/((-8)^2)=1 is Ellipse with (h,k)=(-2,3),a=13,b=8