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

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

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

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

(h, k) = (2, - 1), a = 4, b = 5

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (2, - 1), a = 4, b = 5

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (2, - 1), b = 5, a = 4

\frac{x ^{2}}{64} + \frac{y ^{2}}{25} = 1

3 x^{2} + 4 y^{2} - 16 x + 16 = 0

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

(6 (x)^{2} - 12 x) + ((y)^{2} + 4 y) = - 4

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

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