What is the foci 1.5+17.67t-4.905t^2 ?

The foci 1.5+17.67t-4.905t^2 is (1.80122…,17.36283…)

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foci 1.5+17.67t-4.905t^2

foci

1.5 + 17.67 t - 4.905 t^{2}

(1.80122 \dots, 17.36283 \dots)

Solution steps

Rewrite

y = 1.5 + 17.67 t - 4.905 t^{2}

in the standard form

4 (- 0.05096 \dots) (y - 17.41380 \dots) = (t - 1.80122 \dots)^{2}

(h, k) = (1.80122 \dots, 17.41380 \dots), p = - 0.05096 \dots

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

p

from the center

(1.80122 \dots, 17.41380 \dots)

along the y-axis

(1.80122 \dots, 17.41380 \dots + p)

= (1.80122 \dots, 17.41380 \dots + (- 0.05096 \dots))

Refine

(1.80122 \dots, 17.36283 \dots)

y = (3 x - 4) (4 x - 3) - 2 x

x^{2} - 4 x - y + 3 = 0

- 2 x^{2} + 24 x + y - 70 = 0

y = - 4 x^{2} + 1

x^{2} - 20 y - 2 x - 39 = 0

What is the foci 1.5+17.67t-4.905t^2 ?
The foci 1.5+17.67t-4.905t^2 is (1.80122…,17.36283…)