What is ((x-2)^2)/(10)+((y+1)^2)/(25)=1 ?

The solution to ((x-2)^2)/(10)+((y+1)^2)/(25)=1 is Ellipse with (h,k)=(2,-1),a=sqrt(10),b=5

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((x-2)^2)/(10)+((y+1)^2)/(25)=1

\frac{( x - 2 ) ^{2}}{10} + \frac{( y + 1 ) ^{2}}{25} = 1

(h, k) = (2, - 1), a = 10, b = 5

Solution steps

\frac{( x - 2 ) ^{2}}{10} + \frac{( y + 1 ) ^{2}}{25} = 1

Rewrite

\frac{( x - 2 ) ^{2}}{10} + \frac{( y + 1 ) ^{2}}{25} = 1

in the form of the standard ellipse equation

\frac{( x - 2 ) ^{2}}{( 10 ) ^{2}} + \frac{( y - ( - 1 ) ) ^{2}}{5 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (2, - 1), a = 10, 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 = 10

36 x^{2} + 16 y^{2} + 144 x - 160 y - 32 = 0

9 x^{2} + 49 y^{2} - 114 x + 490 y + 1360 = 0

x^{2} + 4 z^{2} = 4

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

\frac{( x + 1 ) ^{2}}{9} + \frac{( y - 2 ) ^{2}}{8 ^{2}} = 1

What is ((x-2)^2)/(10)+((y+1)^2)/(25)=1 ?
The solution to ((x-2)^2)/(10)+((y+1)^2)/(25)=1 is Ellipse with (h,k)=(2,-1),a=sqrt(10),b=5