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})

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

y = 8 x^{2}

y = 6 x - x^{2} - 5

(y + 3)^{2} = 4 (x - 3)

y = - 2 x^{2} + 24 x