What is 5^2+5y^2+10x+2y+3=0 ?

The solution to 5^2+5y^2+10x+2y+3=0 is Parabola with (h,k)=(-139/50 ,-1/5),p=-1/2

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5^2+5y^2+10x+2y+3=0

5^{2} + 5 y^{2} + 10 x + 2 y + 3 = 0

(h, k) = (- \frac{139}{50}, - \frac{1}{5}), p = - \frac{1}{2}

Solution steps

5^{2} + 5 y^{2} + 10 x + 2 y + 3 = 0

Rewrite

5^{2} + 5 y^{2} + 10 x + 2 y + 3 = 0

in the standard form

4 (- \frac{1}{2}) (x - (- \frac{139}{50})) = (y - (- \frac{1}{5}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{139}{50}, - \frac{1}{5}), p = - \frac{1}{2}

2 Y = - 2 X^{2}

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

(x - 0)^{2} = 4 (6) (y - 10)

2 x^{2} - 10.4 y + 11.1 = 0

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

What is 5^2+5y^2+10x+2y+3=0 ?
The solution to 5^2+5y^2+10x+2y+3=0 is Parabola with (h,k)=(-139/50 ,-1/5),p=-1/2