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

The solution to ((x+1)^2}{27/4}+\frac{(y-4)^2)/9 =1 is Ellipse with (h,k)=(-1,4),a=(3sqrt(3))/2 ,b=3

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

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

(h, k) = (- 1, 4), a = \frac{3 3}{2}, b = 3

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 1, 4), a = \frac{3 3}{2}, b = 3

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 1, 4), b = 3, a = \frac{3 3}{2}

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

4 x^{2} + 9 y^{2} - 8 x - 54 y - 5 = 0

x^{2} + 4 y^{2} + 8 x + 8 y - 16 = 0

What is ((x+1)^2}{27/4}+\frac{(y-4)^2)/9 =1 ?
The solution to ((x+1)^2}{27/4}+\frac{(y-4)^2)/9 =1 is Ellipse with (h,k)=(-1,4),a=(3sqrt(3))/2 ,b=3