What is ((x-4)^2)/(12)+((y+3)^2)/(16)=1 ?

The solution to ((x-4)^2)/(12)+((y+3)^2)/(16)=1 is Ellipse with (h,k)=(4,-3),a=2sqrt(3),b=4

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((x-4)^2)/(12)+((y+3)^2)/(16)=1

\frac{( x - 4 ) ^{2}}{12} + \frac{( y + 3 ) ^{2}}{16} = 1

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

Solution steps

\frac{( x - 4 ) ^{2}}{12} + \frac{( y + 3 ) ^{2}}{16} = 1

Rewrite

\frac{( x - 4 ) ^{2}}{12} + \frac{( y + 3 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

x^{2} + y^{2} = \frac{2 x}{x ^{2} + y ^{2}}

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

4 x^{2} - 40 x + 9 y^{2} + 54 y + 145 = 0

9 X^{2} + Y^{2} = 81

\frac{( x ^{2} - 2 x + 1 )}{34} + \frac{( y ^{2} - 8 y + 16 )}{25} = 1

What is ((x-4)^2)/(12)+((y+3)^2)/(16)=1 ?
The solution to ((x-4)^2)/(12)+((y+3)^2)/(16)=1 is Ellipse with (h,k)=(4,-3),a=2sqrt(3),b=4