What is ((x-4)^2)/(36)+((y+1)^2)/(10)=1 ?

The solution to ((x-4)^2)/(36)+((y+1)^2)/(10)=1 is Ellipse with (h,k)=(4,-1),a=6,b=sqrt(10)

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((x-4)^2)/(36)+((y+1)^2)/(10)=1

\frac{( x - 4 ) ^{2}}{36} + \frac{( y + 1 ) ^{2}}{10} = 1

(h, k) = (4, - 1), a = 6, b = 10

Solution steps

\frac{( x - 4 ) ^{2}}{36} + \frac{( y + 1 ) ^{2}}{10} = 1

Rewrite

\frac{( x - 4 ) ^{2}}{36} + \frac{( y + 1 ) ^{2}}{10} = 1

in the form of the standard ellipse equation

\frac{( x - 4 ) ^{2}}{6 ^{2}} + \frac{( y - ( - 1 ) ) ^{2}}{( 10 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (4, - 1), a = 6, b = 10

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (4, - 1), a = 6, b = 10

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

\frac{( x - 5 ) ^{2}}{47} + \frac{( y + 3 ) ^{2}}{30} = 1

4 x^{2} - 16 x + 25 y^{2} - 84 = 0

16 (x - 4)^{2} + 25 (y + 2)^{2} = 400

8 x^{2} + 16 y^{2} = 1

What is ((x-4)^2)/(36)+((y+1)^2)/(10)=1 ?
The solution to ((x-4)^2)/(36)+((y+1)^2)/(10)=1 is Ellipse with (h,k)=(4,-1),a=6,b=sqrt(10)