What is ((x-2)^2)/(16)+((y+3)^2)/(25)=1 ?

The solution to ((x-2)^2)/(16)+((y+3)^2)/(25)=1 is Ellipse with (h,k)=(2,-3),a=4,b=5

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((x-2)^2)/(16)+((y+3)^2)/(25)=1

\frac{( x - 2 ) ^{2}}{16} + \frac{( y + 3 ) ^{2}}{25} = 1

(h, k) = (2, - 3), a = 4, b = 5

Solution steps

\frac{( x - 2 ) ^{2}}{16} + \frac{( y + 3 ) ^{2}}{25} = 1

Rewrite

\frac{( x - 2 ) ^{2}}{16} + \frac{( y + 3 ) ^{2}}{25} = 1

in the form of the standard ellipse equation

\frac{( x - 2 ) ^{2}}{4 ^{2}} + \frac{( y - ( - 3 ) ) ^{2}}{5 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (2, - 3), a = 4, b = 5

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (2, - 3), b = 5, a = 4

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

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

1 x^{2} + 3 y^{2} + 9 x + 8 y - 14 = 0

25 x^{2} + 16 y^{2} + 100 x - 64 y - 236 = 0

9 x^{2} + 4 y^{2} - 18 x - 24 y + 9 = 0

What is ((x-2)^2)/(16)+((y+3)^2)/(25)=1 ?
The solution to ((x-2)^2)/(16)+((y+3)^2)/(25)=1 is Ellipse with (h,k)=(2,-3),a=4,b=5