What is the foci y^2-10x+4y-16=0 ?

The foci y^2-10x+4y-16=0 is (1/2 ,-2)

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foci y^2-10x+4y-16=0

foci

y^{2} - 10 x + 4 y - 16 = 0

(\frac{1}{2}, - 2)

Solution steps

Rewrite

y^{2} - 10 x + 4 y - 16 = 0

in the standard form

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

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

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

p

from the center

(- 2, - 2)

along the x-axis

(- 2 + p, - 2)

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

Refine

(\frac{1}{2}, - 2)

x^{2} - 4 x - 5

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

x^{2} + 4 x - 8 y + 12 = 0

y = x^{2} + x + 1

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