What is the foci y^2=5y+7x+3 ?

The foci y^2=5y+7x+3 is (3/7 , 5/2)

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foci y^2=5y+7x+3

foci

y^{2} = 5 y + 7 x + 3

(\frac{3}{7}, \frac{5}{2})

Solution steps

Rewrite

y^{2} = 5 y + 7 x + 3

in the standard form

4 \cdot \frac{7}{4} (x - (- \frac{37}{28})) = (y - \frac{5}{2})^{2}

(h, k) = (- \frac{37}{28}, \frac{5}{2}), p = \frac{7}{4}

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

p

from the center

(- \frac{37}{28}, \frac{5}{2})

along the x-axis

(- \frac{37}{28} + p, \frac{5}{2})

= (- \frac{37}{28} + \frac{7}{4}, \frac{5}{2})

Refine

(\frac{3}{7}, \frac{5}{2})

y = x^{2} - 5 x - 24

x^{2} - 1

y = x^{2} - 4 x

x^{2} - 6 x + 5 y = - 34

5 x (2 x + 4) - 3 x