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

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

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

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

(h, k) = (3, - 1), a = 4, b = 5

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (3, - 1), a = 4, b = 5

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, - 1), b = 5, a = 4

\frac{( x - 3 ) ^{2}}{25} + \frac{( y - 2 ) ^{2}}{9} = 1

4 x^{2} + 9 y^{2} - 24 x + 72 y + 144 = 0

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

x^{2} + 5 y^{2} = 2

x^{2} + 4 y^{2} + 2 x + 8 y + 1 = 0

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