What is (x^2-4)/(6^2)+(y^2+4)/(2^2)=1 ?

The solution to (x^2-4)/(6^2)+(y^2+4)/(2^2)=1 is Ellipse with (h,k)=(0,0),a=2,b= 2/3

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(x^2-4)/(6^2)+(y^2+4)/(2^2)=1

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

(h, k) = (0, 0), a = 2, b = \frac{2}{3}

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (0, 0), a = 2, b = \frac{2}{3}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 0), a = 2, b = \frac{2}{3}

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

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

\frac{( x - 2 ) ^{2}}{16} + \frac{( y - 2 ) ^{2}}{9} = 1

(x + 2)^{2} + 32 (y - 4)^{2} = 16

\frac{( x - 3 ) ^{2}}{16} + \frac{( y - 5 ) ^{2}}{25} = 1

What is (x^2-4)/(6^2)+(y^2+4)/(2^2)=1 ?
The solution to (x^2-4)/(6^2)+(y^2+4)/(2^2)=1 is Ellipse with (h,k)=(0,0),a=2,b= 2/3