What is ((x+3)^2)/4+((y-4)^2)/(18)=1 ?

The solution to ((x+3)^2)/4+((y-4)^2)/(18)=1 is Ellipse with (h,k)=(-3,4),a=2,b=3sqrt(2)

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((x+3)^2)/4+((y-4)^2)/(18)=1

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

(h, k) = (- 3, 4), a = 2, b = 32

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 3, 4), a = 2, b = 32

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 3, 4), b = 32, a = 2

\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

\frac{( x - 5 ) ^{2}}{25} + \frac{( y - 3 ) ^{2}}{9} = 1

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

What is ((x+3)^2)/4+((y-4)^2)/(18)=1 ?
The solution to ((x+3)^2)/4+((y-4)^2)/(18)=1 is Ellipse with (h,k)=(-3,4),a=2,b=3sqrt(2)