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

The solution to (3x^2)/4+((4y)/4)^2= 12/12 is Ellipse with (h,k)=(0,0),a= 2/(sqrt(3)),b=1

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

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

(h, k) = (0, 0), a = \frac{2}{3}, b = 1

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (0, 0), a = \frac{2}{3}, b = 1

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 0), a = \frac{2}{3}, b = 1

9 x^{2} + 25 y^{2} - 18 x - 250 y + 409 = 0

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

a - bi = \frac{1}{a + bi}

x^{2} + 4 y^{2} + 20 x + 44 y + 121 = 0

16 x^{2} + 7 y^{2} = 112

What is (3x^2)/4+((4y)/4)^2= 12/12 ?
The solution to (3x^2)/4+((4y)/4)^2= 12/12 is Ellipse with (h,k)=(0,0),a= 2/(sqrt(3)),b=1