What is ((x-3)^2)/(64)+((y+1)^2)/(25)=1 ?

The solution to ((x-3)^2)/(64)+((y+1)^2)/(25)=1 is Ellipse with (h,k)=(3,-1),a=8,b=5

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((x-3)^2)/(64)+((y+1)^2)/(25)=1

\frac{( x - 3 ) ^{2}}{64} + \frac{( y + 1 ) ^{2}}{25} = 1

(h, k) = (3, - 1), a = 8, b = 5

Solution steps

\frac{( x - 3 ) ^{2}}{64} + \frac{( y + 1 ) ^{2}}{25} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{64} + \frac{( y + 1 ) ^{2}}{25} = 1

in the form of the standard ellipse equation

\frac{( x - 3 ) ^{2}}{8 ^{2}} + \frac{( y - ( - 1 ) ) ^{2}}{5 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (3, - 1), a = 8, b = 5

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (3, - 1), a = 8, b = 5

\frac{x ^{2}}{25} + \frac{y ^{2}}{9} < 1

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

\frac{x ^{2}}{9} + \frac{y ^{2}}{\frac{9}{25}} = 1

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

3 x^{2} + y^{2} - 12 = 0

What is ((x-3)^2)/(64)+((y+1)^2)/(25)=1 ?
The solution to ((x-3)^2)/(64)+((y+1)^2)/(25)=1 is Ellipse with (h,k)=(3,-1),a=8,b=5