What is the foci y=3x^2-35x ?

The foci y=3x^2-35x is (35/6 ,-102)

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foci y=3x^2-35x

foci

y = 3 x^{2} - 35 x

(\frac{35}{6}, - 102)

Solution steps

Rewrite

y = 3 x^{2} - 35 x

in the standard form

4 \cdot \frac{1}{12} (y - (- \frac{1225}{12})) = (x - \frac{35}{6})^{2}

(h, k) = (\frac{35}{6}, - \frac{1225}{12}), p = \frac{1}{12}

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

p

from the center

(\frac{35}{6}, - \frac{1225}{12})

along the y-axis

(\frac{35}{6}, - \frac{1225}{12} + p)

= (\frac{35}{6}, - \frac{1225}{12} + \frac{1}{12})

Refine

(\frac{35}{6}, - 102)

y^{2} - 6 y - 2 x + 7 = 0

y = - 0.0000848 x^{2} - 0.006005 x + 1.004

y^{2} - \frac{3}{4} x = 0

40 x - 5 x^{2}

x = - \frac{3}{8} (y - 2)^{2} - 1