What is the foci x^2+2x-1-(2x^2-3x+6) ?

The foci x^2+2x-1-(2x^2-3x+6) is (5/2 ,-1)

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foci x^2+2x-1-(2x^2-3x+6)

foci

x^{2} + 2 x - 1 - (2 x^{2} - 3 x + 6)

(\frac{5}{2}, - 1)

Solution steps

Rewrite

y = x^{2} + 2 x - 1 - (2 x^{2} - 3 x + 6)

in the standard form

4 (- \frac{1}{4}) (y - (- \frac{3}{4})) = (x - \frac{5}{2})^{2}

(h, k) = (\frac{5}{2}, - \frac{3}{4}), p = - \frac{1}{4}

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

p

from the center

(\frac{5}{2}, - \frac{3}{4})

along the y-axis

(\frac{5}{2}, - \frac{3}{4} + p)

= (\frac{5}{2}, - \frac{3}{4} + (- \frac{1}{4}))

Refine

(\frac{5}{2}, - 1)

- x^{2} + 8 x - 12

y = x^{2} + 5 x - 14

y^{2} - 10 y - 12 x + 13 = 0

y^{2} - 4 x = 0

y^{2} = 5 x

What is the foci x^2+2x-1-(2x^2-3x+6) ?
The foci x^2+2x-1-(2x^2-3x+6) is (5/2 ,-1)