What is ((x-4)^2)/3+(y^2)/9 =1 ?

The solution to ((x-4)^2)/3+(y^2)/9 =1 is Ellipse with (h,k)=(4,0),a=sqrt(3),b=3

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((x-4)^2)/3+(y^2)/9 =1

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

(h, k) = (4, 0), a = 3, b = 3

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

\frac{( x - 4 ) ^{2}}{( 3 ) ^{2}} + \frac{( y - 0 ) ^{2}}{3 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (4, 0), a = 3, b = 3

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (4, 0), b = 3, a = 3

25 x^{2} + 16 y^{2} - 100 x - 96 y - 156 = 0

\frac{x ^{2}}{25} + \frac{( y - 3 ) ^{2}}{16} = 1

\frac{( x + 2 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{4} = 1

16 - x^{2} - 4 y^{2} > 0

9 X^{2} + 4 Y^{2} - 24 Y = 0

What is ((x-4)^2)/3+(y^2)/9 =1 ?
The solution to ((x-4)^2)/3+(y^2)/9 =1 is Ellipse with (h,k)=(4,0),a=sqrt(3),b=3