What is the foci-2x^2+20x-y-52=0 ?

The foci-2x^2+20x-y-52=0 is (5,-17/8)

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foci-2x^2+20x-y-52=0

foci

- 2 x^{2} + 20 x - y - 52 = 0

(5, - \frac{17}{8})

Solution steps

Rewrite

- 2 x^{2} + 20 x - y - 52 = 0

in the standard form

4 (- \frac{1}{8}) (y - (- 2)) = (x - 5)^{2}

(h, k) = (5, - 2), p = - \frac{1}{8}

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

p

from the center

(5, - 2)

along the y-axis

(5, - 2 + p)

= (5, - 2 + (- \frac{1}{8}))

Refine

(5, - \frac{17}{8})

(\frac{x - 5}{8})^{2} - (\frac{y + 2}{7}) = 1

5 x^{2}

x^{2} + 20 y = 0

- 0.3 x^{2} + 2 x + 3

6 x^{2} - 3 x 5