What is the foci y=x^2+7x-1 ?

The foci y=x^2+7x-1 is (-7/2 ,-13)

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foci y=x^2+7x-1

foci

y = x^{2} + 7 x - 1

(- \frac{7}{2}, - 13)

Solution steps

Rewrite

y = x^{2} + 7 x - 1

in the standard form

4 \cdot \frac{1}{4} (y - (- \frac{53}{4})) = (x - (- \frac{7}{2}))^{2}

(h, k) = (- \frac{7}{2}, - \frac{53}{4}), p = \frac{1}{4}

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

p

from the center

(- \frac{7}{2}, - \frac{53}{4})

along the y-axis

(- \frac{7}{2}, - \frac{53}{4} + p)

= (- \frac{7}{2}, - \frac{53}{4} + \frac{1}{4})

Refine

(- \frac{7}{2}, - 13)

+ 3 y^{2} = 24 x

y = \frac{18 x ^{6}}{27 x ^{4}}

y^{2} - 2 y - 8 x - 39 = 0

- 2 x^{2} + 20 x - y - 52 = 0

(\frac{x - 5}{8})^{2} - (\frac{y + 2}{7}) = 1