What is the foci y=(2X+5)(3X-7) ?

The foci y=(2X+5)(3X-7) is (-35,-1/12)

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foci y=(2X+5)(3X-7)

foci

y = (2 X + 5) (3 X - 7)

(- 35, - \frac{1}{12})

Solution steps

Rewrite

y = (2 X + 5) (3 X - 7)

in the standard form

4 \cdot \frac{1}{24} (y - (- \frac{841}{24})) = (X - (- \frac{1}{12}))^{2}

(h, k) = (- \frac{841}{24}, - \frac{1}{12}), p = \frac{1}{24}

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

p

from the center

(- \frac{841}{24}, - \frac{1}{12})

along the x-axis

(- \frac{841}{24} + p, - \frac{1}{12})

= (- \frac{841}{24} + \frac{1}{24}, - \frac{1}{12})

Refine

(- 35, - \frac{1}{12})

y = \frac{1}{2} x^{2}

y^{2} = 5 y + 7 x + 3

y = x^{2} - 5 x - 24

x^{2} - 1

y = x^{2} - 4 x