What is 4x^2+6y^2-8x+4y-6>= 0 ?

The solution to 4x^2+6y^2-8x+4y-6>= 0 is Ellipse with (h,k)=(1,-1/3),a=(2sqrt(2))/(sqrt(3)),b= 4/3

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4x^2+6y^2-8x+4y-6>= 0

4 x^{2} + 6 y^{2} - 8 x + 4 y - 6 \geq 0

(h, k) = (1, - \frac{1}{3}), a = \frac{2 2}{3}, b = \frac{4}{3}

Solution steps

4 x^{2} + 6 y^{2} - 8 x + 4 y - 6 \geq 0

Rewrite

4 x^{2} + 6 y^{2} - 8 x + 4 y - 6 \geq 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (1, - \frac{1}{3}), a = \frac{2 2}{3}, b = \frac{4}{3}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (1, - \frac{1}{3}), a = \frac{2 2}{3}, b = \frac{4}{3}

25 x^{2} + 4 y^{2} + 200 x - 8 y + 304 = 0

(x - 3)^{2} + 16 (y - 2)^{2} = 16

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

\frac{( x + 1 ) ^{2}}{100} + \frac{( y + 2 ) ^{2}}{121} = 1

25 x^{2} + 16 y^{2} < 400

What is 4x^2+6y^2-8x+4y-6>= 0 ?
The solution to 4x^2+6y^2-8x+4y-6>= 0 is Ellipse with (h,k)=(1,-1/3),a=(2sqrt(2))/(sqrt(3)),b= 4/3