What is ((x-3)^2)/(4^2)+((y+2)^2)/(3^2)=1 ?

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

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((x-3)^2)/(4^2)+((y+2)^2)/(3^2)=1

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

(h, k) = (3, - 2), a = 4, b = 3

Solution steps

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

Therefore ellipse properties are:

(h, k) = (3, - 2), a = 4, b = 3

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (3, - 2), a = 4, b = 3

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

\frac{x ^{2}}{256} + \frac{y ^{2}}{400} = 2

3 x^{2} + 4 y^{2} + 6 x - 24 y + 27 = 0

25 x^{2} + 9 y^{2} - 100 x + 18 x - 10 y - 31 = 0

- X^{2} - 4 = Y^{2}

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