What is the foci 4x^2-12x-16y+41=0 ?

The foci 4x^2-12x-16y+41=0 is (3/2 ,3)

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foci 4x^2-12x-16y+41=0

foci

4 x^{2} - 12 x - 16 y + 41 = 0

(\frac{3}{2}, 3)

Solution steps

Rewrite

4 x^{2} - 12 x - 16 y + 41 = 0

in the standard form

4 \cdot 1 \cdot (y - 2) = (x - \frac{3}{2})^{2}

(h, k) = (\frac{3}{2}, 2), p = 1

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

p

from the center

(\frac{3}{2}, 2)

along the y-axis

(\frac{3}{2}, 2 + p)

= (\frac{3}{2}, 2 + 1)

Refine

(\frac{3}{2}, 3)

1 - x - x^{2}

4 x^{2} - 20 x - 24 y + 97 = 0

y^{2} = x + 1

y = x^{2} - 4 x 3

(x + 2) (x - 2)