What is the directrix y^2=(-16)/5 x ?

The directrix y^2=(-16)/5 x is x= 4/5

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directrix y^2=(-16)/5 x

directrix

y^{2} = \frac{- 16}{5} x

x = \frac{4}{5}

Solution steps

Rewrite

y^{2} = \frac{- 16}{5} x

in the standard form:

4 (- \frac{4}{5}) (x - 0) = (y - 0)^{2}

(h, k) = (0, 0), p = - \frac{4}{5}

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

- p

from the center

(0, 0)

x-coordinate

x = 0 - p

x = 0 - (- \frac{4}{5})

Refine

x = \frac{4}{5}

y = (2 X + 5) (3 X - 7)

(y + 2)^{2} = - 16 (x + 9)

x^{2} + 5 x - 84

2 x^{2} - 32 x - y + 127 = 0

y^{2} + 2 x + 6 y + 17 = 0