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

The foci y^2=5x+3 is (13/20 ,0)

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foci y^2=5x+3

foci

y^{2} = 5 x + 3

(\frac{13}{20}, 0)

Solution steps

Rewrite

y^{2} = 5 x + 3

in the standard form:

4 \cdot \frac{5}{4} (x - (- \frac{3}{5})) = (y - 0)^{2}

(h, k) = (- \frac{3}{5}, 0), p = \frac{5}{4}

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

p

from the center

(- \frac{3}{5}, 0)

along the x-axis

(- \frac{3}{5} + p, 0)

= (- \frac{3}{5} + \frac{5}{4}, 0)

Refine

(\frac{13}{20}, 0)

(x + 3)^{2} = - 2 (y - 1)

y = (x + \frac{1}{4}) (x - 5)

y^{2} - 16 x = 0

6 x^{2}

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