What is the foci (x-5)(x+2) ?

The foci (x-5)(x+2) is (3/2 ,-12)

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foci (x-5)(x+2)

foci

(x - 5) (x + 2)

(\frac{3}{2}, - 12)

Solution steps

Rewrite

y = (x - 5) (x + 2)

in the standard form

4 \cdot \frac{1}{4} (y - (- \frac{49}{4})) = (x - \frac{3}{2})^{2}

(h, k) = (\frac{3}{2}, - \frac{49}{4}), p = \frac{1}{4}

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

p

from the center

(\frac{3}{2}, - \frac{49}{4})

along the y-axis

(\frac{3}{2}, - \frac{49}{4} + p)

= (\frac{3}{2}, - \frac{49}{4} + \frac{1}{4})

Refine

(\frac{3}{2}, - 12)

(x + 1)^{2} + 3

y^{2} + 2 y + 12 x + 25 = 0

(x - 3)^{2} = y - 7

- 3 x^{2} + 5 x

(x - 1)^{2} = - 2 y - 11