What is ((x-4)^2)/4+((y+2)^2)/(20)=1 ?

The solution to ((x-4)^2)/4+((y+2)^2)/(20)=1 is Ellipse with (h,k)=(4,-2),a=2,b=2sqrt(5)

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((x-4)^2)/4+((y+2)^2)/(20)=1

\frac{( x - 4 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{20} = 1

(h, k) = (4, - 2), a = 2, b = 25

Solution steps

\frac{( x - 4 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{20} = 1

Rewrite

\frac{( x - 4 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{20} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (4, - 2), a = 2, b = 25

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (4, - 2), b = 25, a = 2

\frac{x ^{2}}{13} + \frac{y ^{2}}{9} = 1

\frac{( x + 3 ) ^{2}}{4} + \frac{( y - 4 ) ^{2}}{18} = 1

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

\frac{( x + 1 ) ^{2}}{\frac{27}{4}} + \frac{( y - 4 ) ^{2}}{9} = 1

9 - x^{2} - 9 y^{2} = 0

What is ((x-4)^2)/4+((y+2)^2)/(20)=1 ?
The solution to ((x-4)^2)/4+((y+2)^2)/(20)=1 is Ellipse with (h,k)=(4,-2),a=2,b=2sqrt(5)