What is the foci 96t-16t^2 ?

The foci 96t-16t^2 is (3, 9215/64)

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foci 96t-16t^2

foci

96 t - 16 t^{2}

(3, \frac{9215}{64})

Solution steps

Rewrite

y = 96 t - 16 t^{2}

in the standard form

4 (- \frac{1}{64}) (y - 144) = (t - 3)^{2}

(h, k) = (3, 144), p = - \frac{1}{64}

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

p

from the center

(3, 144)

along the y-axis

(3, 144 + p)

= (3, 144 + (- \frac{1}{64}))

Refine

(3, \frac{9215}{64})

y = - 2 x^{2} + x - 1

4 x^{2} + 2 x + 4

y^{2} + 16 x = 0

y = 16 - x^{2}

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