What is the directrix y=(x^4-1)/(x^2-1) ?

The directrix y=(x^4-1)/(x^2-1) is y= 3/4

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directrix y=(x^4-1)/(x^2-1)

directrix

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

y = \frac{3}{4}

Solution steps

Rewrite

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

in the standard form

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

(h, k) = (0, 1), p = \frac{1}{4}

Parabola is symmetric around the y-axis and so the directrix is a line parallel to the x-axis, a distance

- p

from the center

(0, 1)

y-coordinate

y = 1 - p

y = 1 - \frac{1}{4}

Refine

y = \frac{3}{4}

(x - 2) (x + 1)

5 x^{2} + 30 x + 24 y = 51

x^{2} - 2 x - 4 y + 9 = 0

\frac{- 3 x ^{2} - 2 x + 6}{5}

3 x^{2} - x - 2

What is the directrix y=(x^4-1)/(x^2-1) ?
The directrix y=(x^4-1)/(x^2-1) is y= 3/4