What is the foci-x^2+3x=2y+3 ?

The foci-x^2+3x=2y+3 is (3/2 ,-7/8)

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foci-x^2+3x=2y+3

foci

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

(\frac{3}{2}, - \frac{7}{8})

Solution steps

Rewrite

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

in the standard form

4 (- \frac{1}{2}) (y - (- \frac{3}{8})) = (x - \frac{3}{2})^{2}

(h, k) = (\frac{3}{2}, - \frac{3}{8}), p = - \frac{1}{2}

Parabola is symmetric around the y-axis and so the focus lies a distance

p

from the center

(\frac{3}{2}, - \frac{3}{8})

along the y-axis

(\frac{3}{2}, - \frac{3}{8} + p)

= (\frac{3}{2}, - \frac{3}{8} + (- \frac{1}{2}))

Refine

(\frac{3}{2}, - \frac{7}{8})

- x^{2} + 10 x - 23

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

x^{2} - 5 x + 6

(x + 2)^{2} = - 6 (y - 1)

y = + x^{2} + 81