What is ((x-3)^2)/(64)+((y+4)^2)/(81)=1 ?

The solution to ((x-3)^2)/(64)+((y+4)^2)/(81)=1 is Ellipse with (h,k)=(3,-4),a=8,b=9

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((x-3)^2)/(64)+((y+4)^2)/(81)=1

\frac{( x - 3 ) ^{2}}{64} + \frac{( y + 4 ) ^{2}}{81} = 1

(h, k) = (3, - 4), a = 8, b = 9

Solution steps

\frac{( x - 3 ) ^{2}}{64} + \frac{( y + 4 ) ^{2}}{81} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{64} + \frac{( y + 4 ) ^{2}}{81} = 1

in the form of the standard ellipse equation

\frac{( x - 3 ) ^{2}}{8 ^{2}} + \frac{( y - ( - 4 ) ) ^{2}}{9 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (3, - 4), a = 8, b = 9

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, - 4), b = 9, a = 8

49 x^{2} + 4 y^{2} + 98 x - 147 = 0

9 x^{2} + 25 y^{2} - 144 x + 200 y + 751 = 0

\frac{4 x ^{2}}{19} + \frac{2 y ^{2}}{19} = 1

2 x - x^{2} - 4 y^{2} - 16 y - 1 = 0

\frac{( x - 4 ) ^{2}}{16} + \frac{y ^{2}}{40.96} = 1

What is ((x-3)^2)/(64)+((y+4)^2)/(81)=1 ?
The solution to ((x-3)^2)/(64)+((y+4)^2)/(81)=1 is Ellipse with (h,k)=(3,-4),a=8,b=9