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

The foci (2x-5)^2 is (5/2 , 1/16)

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foci (2x-5)^2

foci

(2 x - 5)^{2}

(\frac{5}{2}, \frac{1}{16})

Solution steps

Rewrite

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

in the standard form:

4 \cdot \frac{1}{16} (y - 0) = (x - \frac{5}{2})^{2}

(h, k) = (\frac{5}{2}, 0), p = \frac{1}{16}

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

p

from the center

(\frac{5}{2}, 0)

along the y-axis

(\frac{5}{2}, 0 + p)

= (\frac{5}{2}, 0 + \frac{1}{16})

Refine

(\frac{5}{2}, \frac{1}{16})

y = + x^{2} + 6 x + 3

4 (x + 2)^{2} + 5

(y - 8)^{2} = 16 (x - 4)

y = \frac{x ^{3} - 3 x ^{2} - 4 x}{x - 4}

y = s^{2}