What is the foci y=2x^2-7x-3 ?

The foci y=2x^2-7x-3 is (7/4 ,-9)

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foci y=2x^2-7x-3

foci

y = 2 x^{2} - 7 x - 3

(\frac{7}{4}, - 9)

Solution steps

Rewrite

y = 2 x^{2} - 7 x - 3

in the standard form

4 \cdot \frac{1}{8} (y - (- \frac{73}{8})) = (x - \frac{7}{4})^{2}

(h, k) = (\frac{7}{4}, - \frac{73}{8}), p = \frac{1}{8}

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

p

from the center

(\frac{7}{4}, - \frac{73}{8})

along the y-axis

(\frac{7}{4}, - \frac{73}{8} + p)

= (\frac{7}{4}, - \frac{73}{8} + \frac{1}{8})

Refine

(\frac{7}{4}, - 9)

12 x^{2} - 72 x + 77

y^{2} - 10 x + 4 y - 16 = 0

x^{2} - 4 x - 5

y = - 2 (x - 3)^{2} + 1

x^{2} + 4 x - 8 y + 12 = 0