What is ((x-5)^2)/(8^2)+((y+2)^2)/(10^2)=1 ?

The solution to ((x-5)^2)/(8^2)+((y+2)^2)/(10^2)=1 is Ellipse with (h,k)=(5,-2),a=8,b=10

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((x-5)^2)/(8^2)+((y+2)^2)/(10^2)=1

\frac{( x - 5 ) ^{2}}{8 ^{2}} + \frac{( y + 2 ) ^{2}}{1 0 ^{2}} = 1

(h, k) = (5, - 2), a = 8, b = 10

Solution steps

\frac{( x - 5 ) ^{2}}{8 ^{2}} + \frac{( y + 2 ) ^{2}}{1 0 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (5, - 2), a = 8, b = 10

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (5, - 2), b = 10, a = 8

2 y - x^{2} - y^{2} \geq 0

25 x^{2} + 4 y^{2} - 8 x + 200 y + 304 = 0

2 x^{2} + y^{2} + 2 y = 1

\frac{x ^{2}}{25} + \frac{( y - 2 ) ^{2}}{4} = 1

y - 3 = - \frac{x}{4 y} (x - 12)

What is ((x-5)^2)/(8^2)+((y+2)^2)/(10^2)=1 ?
The solution to ((x-5)^2)/(8^2)+((y+2)^2)/(10^2)=1 is Ellipse with (h,k)=(5,-2),a=8,b=10