What is the foci+(5x-1)^2 ?

The foci+(5x-1)^2 is (1/5 , 1/100)

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foci+(5x-1)^2

foci

+ (5 x - 1)^{2}

(\frac{1}{5}, \frac{1}{100})

Solution steps

Rewrite

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

in the standard form:

4 \cdot \frac{1}{100} (y - 0) = (x - \frac{1}{5})^{2}

(h, k) = (\frac{1}{5}, 0), p = \frac{1}{100}

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

p

from the center

(\frac{1}{5}, 0)

along the y-axis

(\frac{1}{5}, 0 + p)

= (\frac{1}{5}, 0 + \frac{1}{100})

Refine

(\frac{1}{5}, \frac{1}{100})

y = x^{2} + 2 x + 12

x = (y + 4)^{2}

y = 6 x^{2} - 1

1.5 + 17.67 t - 4.905 t^{2}

y = (3 x - 4) (4 x - 3) - 2 x