What is 3x^2+4y^2+28x-16y+48=0 ?

The solution to 3x^2+4y^2+28x-16y+48=0 is Ellipse with (h,k)=(-14/3 ,2),a= 10/3 ,b= 5/(sqrt(3))

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3x^2+4y^2+28x-16y+48=0

3 x^{2} + 4 y^{2} + 28 x - 16 y + 48 = 0

(h, k) = (- \frac{14}{3}, 2), a = \frac{10}{3}, b = \frac{5}{3}

Solution steps

3 x^{2} + 4 y^{2} + 28 x - 16 y + 48 = 0

Rewrite

3 x^{2} + 4 y^{2} + 28 x - 16 y + 48 = 0

in the form of the standard ellipse equation

\frac{( x - ( - \frac{14}{3} ) ) ^{2}}{( \frac{10}{3} ) ^{2}} + \frac{( y - 2 ) ^{2}}{( \frac{5}{3} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- \frac{14}{3}, 2), a = \frac{10}{3}, b = \frac{5}{3}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- \frac{14}{3}, 2), a = \frac{10}{3}, b = \frac{5}{3}

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5 x^{2} + 9 y^{2} + 30 x = 0

4 x^{2} + 7 y^{2} = 28

5 u^{2} - 4 \cdot u + v^{2} + 2 \cdot v = 0

What is 3x^2+4y^2+28x-16y+48=0 ?
The solution to 3x^2+4y^2+28x-16y+48=0 is Ellipse with (h,k)=(-14/3 ,2),a= 10/3 ,b= 5/(sqrt(3))