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Popular >

foci 4t^2+1

Solution

foci

4 t^{2} + 1

Solution

(0, \frac{17}{16})

Solution steps

Rewrite

y = 4 t^{2} + 1

in the standard form:

4 \cdot \frac{1}{16} (y - 1) = (t - 0)^{2}

(h, k) = (0, 1), p = \frac{1}{16}

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

p

from the center

(0, 1)

along the y-axis

(0, 1 + p)

= (0, 1 + \frac{1}{16})

Refine

(0, \frac{17}{16})

Popular Examples

foci t^2+2t-48 foci

t^{2} + 2 t - 48

foci x=-1/2 (y-4)^2+1/2 foci

x = - \frac{1}{2} (y - 4)^{2} + \frac{1}{2}

foci x^2+8x-4y-36=0 foci

x^{2} + 8 x - 4 y - 36 = 0

foci 4x^2+8x+6 foci

4 x^{2} + 8 x + 6

foci y=(2x-1)(5x-6)foci

y = (2 x - 1) (5 x - 6)

Frequently Asked Questions (FAQ)

What is the foci 4t^2+1 ?
The foci 4t^2+1 is (0, 17/16)