What is the foci y=196x^2-112x-240 ?

The foci y=196x^2-112x-240 is (2/7 ,-200703/784)

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foci y=196x^2-112x-240

foci

y = 196 x^{2} - 112 x - 240

(\frac{2}{7}, - \frac{200703}{784})

Solution steps

Rewrite

y = 196 x^{2} - 112 x - 240

in the standard form

4 \cdot \frac{1}{784} (y - (- 256)) = (x - \frac{2}{7})^{2}

(h, k) = (\frac{2}{7}, - 256), p = \frac{1}{784}

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

p

from the center

(\frac{2}{7}, - 256)

along the y-axis

(\frac{2}{7}, - 256 + p)

= (\frac{2}{7}, - 256 + \frac{1}{784})

Refine

(\frac{2}{7}, - \frac{200703}{784})

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

y = x^{2} - 5 x 3

x^{2} - 3 x + 4

x = \frac{( y - 3 ) ^{2}}{8} + 5

y = - 2 x^{2}

What is the foci y=196x^2-112x-240 ?
The foci y=196x^2-112x-240 is (2/7 ,-200703/784)