What is 31x^2=1080y-558 ?

The solution to 31x^2=1080y-558 is Parabola with (h,k)=(0, 31/60),p= 270/31

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31x^2=1080y-558

31 x^{2} = 1080 y - 558

(h, k) = (0, \frac{31}{60}), p = \frac{270}{31}

Solution steps

31 x^{2} = 1080 y - 558

Rewrite

31 x^{2} = 1080 y - 558

in the standard form:

4 \cdot \frac{270}{31} (y - \frac{31}{60}) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, \frac{31}{60}), p = \frac{270}{31}

x = 18 - y^{2}

y^{2} + y + x - 20 = 0.2 y^{2} - x - 3 y - 4

x^{2} + 6 x + 16 y - 7 = 0

2 x^{2} + 8 y - 12 = 0

- 5 x^{2} + 7 y - 11 = 0

What is 31x^2=1080y-558 ?
The solution to 31x^2=1080y-558 is Parabola with (h,k)=(0, 31/60),p= 270/31