What is the foci y^2-10y+4x+21=0 ?

The foci y^2-10y+4x+21=0 is (0,5)

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foci y^2-10y+4x+21=0

foci

y^{2} - 10 y + 4 x + 21 = 0

(0, 5)

Solution steps

Rewrite

y^{2} - 10 y + 4 x + 21 = 0

in the standard form

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

(h, k) = (1, 5), p = - 1

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

p

from the center

(1, 5)

along the x-axis

(1 + p, 5)

= (1 + (- 1), 5)

Refine

(0, 5)

0.3 x^{2} - 66 x + 23114

\frac{x}{25} - y^{2} = 1

y = 2 x^{2} - 2

x^{2} + 2 x - 1 - (2 x^{2} - 3 x + 6)

- x^{2} + 8 x - 12