What is ((x+2)^2}{64}+\frac{(y-3)^2)/b =1 ?

The solution to ((x+2)^2}{64}+\frac{(y-3)^2)/b =1 is Ellipse with (h,k)=(-2,3),a=8,b=sqrt(b)

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((x+2)^2}{64}+\frac{(y-3)^2)/b =1

\frac{( x + 2 ) ^{2}}{64} + \frac{( y - 3 ) ^{2}}{b} = 1

(h, k) = (- 2, 3), a = 8, b = b

Solution steps

\frac{( x + 2 ) ^{2}}{64} + \frac{( y - 3 ) ^{2}}{b} = 1

Rewrite

\frac{( x + 2 ) ^{2}}{64} + \frac{( y - 3 ) ^{2}}{b} = 1

in the form of the standard ellipse equation

\frac{( x - ( - 2 ) ) ^{2}}{8 ^{2}} + \frac{( y - 3 ) ^{2}}{( b ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 2, 3), a = 8, b = b

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 2, 3), a = 8, b = b

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

\frac{( x ^{(2)} )}{( 64 )} + \frac{( y ^{(2)} )}{( 36 )} = 1

\frac{( x + 1 ) ^{2}}{\frac{15}{4}} + \frac{( y + \frac{3}{2} ) ^{2}}{16} = 1

\frac{x ^{2}}{\frac{1}{2}} + \frac{y ^{2}}{\frac{1}{2} - 16} = 1

4 (x - 1)^{2} + 9 (y - 2)^{2} = 36

What is ((x+2)^2}{64}+\frac{(y-3)^2)/b =1 ?
The solution to ((x+2)^2}{64}+\frac{(y-3)^2)/b =1 is Ellipse with (h,k)=(-2,3),a=8,b=sqrt(b)