What is ((x-3)^2)/(12)+((y+4)^2)/(36)=1 ?

The solution to ((x-3)^2)/(12)+((y+4)^2)/(36)=1 is Ellipse with (h,k)=(3,-4),a=2sqrt(3),b=6

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((x-3)^2)/(12)+((y+4)^2)/(36)=1

\frac{( x - 3 ) ^{2}}{12} + \frac{( y + 4 ) ^{2}}{36} = 1

(h, k) = (3, - 4), a = 23, b = 6

Solution steps

\frac{( x - 3 ) ^{2}}{12} + \frac{( y + 4 ) ^{2}}{36} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{12} + \frac{( y + 4 ) ^{2}}{36} = 1

in the form of the standard ellipse equation

\frac{( x - 3 ) ^{2}}{( 2 3 ) ^{2}} + \frac{( y - ( - 4 ) ) ^{2}}{6 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (3, - 4), a = 23, b = 6

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (3, - 4), b = 6, a = 23

1 - \frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 0

4 x^{2} + y^{2} + 16 x + 16 y - 64 = 0

5 x^{2} + 9 y^{2} - 10 x - 36 y = 4

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

\frac{( x + 2 ) ^{2}}{15} + \frac{y ^{2}}{64} = 1

What is ((x-3)^2)/(12)+((y+4)^2)/(36)=1 ?
The solution to ((x-3)^2)/(12)+((y+4)^2)/(36)=1 is Ellipse with (h,k)=(3,-4),a=2sqrt(3),b=6