What is the foci (y+1.5)^2=4(x-2) ?

The foci (y+1.5)^2=4(x-2) is (3,-1.5)

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foci (y+1.5)^2=4(x-2)

foci

(y + 1.5)^{2} = 4 (x - 2)

(3, - 1.5)

Solution steps

Rewrite

(y + 1.5)^{2} = 4 (x - 2)

in the standard form:

4 \cdot 1 \cdot (x - 2) = (y - (- 1.5))^{2}

(h, k) = (2, - 1.5), p = 1

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

p

from the center

(2, - 1.5)

along the x-axis

(2 + p, - 1.5)

= (2 + 1, - 1.5)

Refine

(3, - 1.5)

\frac{1}{2} x^{2} - 7 x + 7

(2 x - 5)^{2}

y = + x^{2} + 6 x + 3

4 (x + 2)^{2} + 5

(y - 8)^{2} = 16 (x - 4)