What is 16x^2+56x+144y+517=0 ?

The solution to 16x^2+56x+144y+517=0 is Parabola with (h,k)=(-7/4 ,-13/4),p=-9/4

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16x^2+56x+144y+517=0

16 x^{2} + 56 x + 144 y + 517 = 0

(h, k) = (- \frac{7}{4}, - \frac{13}{4}), p = - \frac{9}{4}

Solution steps

16 x^{2} + 56 x + 144 y + 517 = 0

Rewrite

16 x^{2} + 56 x + 144 y + 517 = 0

in the standard form

4 (- \frac{9}{4}) (y - (- \frac{13}{4})) = (x - (- \frac{7}{4}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{7}{4}, - \frac{13}{4}), p = - \frac{9}{4}

15 x + 3 y = 7 - 5 x + 4 \cdot 9 y^{2}

3 x^{2} + 5 y - 6 = 0

x^{2} - 4 x - 4 y - 32 = 0

X^{2} + 2 Y = 6

(\frac{1}{6}) (x^{2} + 3 x - 4) = p (x) 6

What is 16x^2+56x+144y+517=0 ?
The solution to 16x^2+56x+144y+517=0 is Parabola with (h,k)=(-7/4 ,-13/4),p=-9/4