What is the foci y=(6x)/7 (4x)/3 ?

The foci y=(6x)/7 (4x)/3 is (0, 7/32)

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foci y=(6x)/7 (4x)/3

foci

y = \frac{6 x}{7} \frac{4 x}{3}

(0, \frac{7}{32})

Solution steps

Rewrite

y = \frac{6 x}{7} \cdot \frac{4 x}{3}

in the standard form

4 \cdot \frac{7}{32} (y - 0) = (x - 0)^{2}

(h, k) = (0, 0), p = \frac{7}{32}

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

p

from the center

(0, 0)

along the y-axis

(0, 0 + p)

= (0, 0 + \frac{7}{32})

Refine

(0, \frac{7}{32})

y = 4 x^{2}

(x - 2) = 3 (y - 5)^{2}

y = 6 x^{2} - 3 x 5

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

- 0.5 x^{2} - 3