What is 5x^2+4x-10y+39=0 ?

The solution to 5x^2+4x-10y+39=0 is Parabola with (h,k)=(-2/5 , 191/50),p= 1/2

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5x^2+4x-10y+39=0

5 x^{2} + 4 x - 10 y + 39 = 0

(h, k) = (- \frac{2}{5}, \frac{191}{50}), p = \frac{1}{2}

Solution steps

5 x^{2} + 4 x - 10 y + 39 = 0

Rewrite

5 x^{2} + 4 x - 10 y + 39 = 0

in the standard form

4 \cdot \frac{1}{2} (y - \frac{191}{50}) = (x - (- \frac{2}{5}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{2}{5}, \frac{191}{50}), p = \frac{1}{2}

2 x + 5 y^{2} = 0

- 3 x - y^{2} = - 1

5 x^{2} - 18 y + 180 = 0

2 x^{2} + y = 75

(x - 5)^{2} = - 70 (y - 1)

What is 5x^2+4x-10y+39=0 ?
The solution to 5x^2+4x-10y+39=0 is Parabola with (h,k)=(-2/5 , 191/50),p= 1/2