can-the-directrix-of-a-parabola-be-in-the-form-y-mx-b-or-is-there-an-inclined-parabola-with-directrix-and-axis-of-symmetry-in-the-form-of-y-mx-b-

Question Number 48121 by JDlix last updated on 19/Nov/18

c a n t h e d i r e c t r i x o f a p a r a b o l a b e i n t h e f o r m y = m x + b ?

o r i s t h e r e a n i n c l i n e d p a r a b o l a w i t h d i r e c t r i x a n d a x i s

o f s y m m e t r y i n t h e f o r m o f y = m x + b ? ?

Commented by MJS last updated on 19/Nov/18

you can rotate a parabola, so this is indeed

possible

Commented by JDlix last updated on 19/Nov/18

c a n i h a v e e x a m p l e ?

Commented by MJS last updated on 20/Nov/18

y = \sqrt{3} x + \frac{13 \sqrt{3}}{8} \pm 2 \sqrt{2 x + 4}

is a parabola with axis y = \sqrt{3} x + \frac{5 \sqrt{3}}{8}

its equation is

192 x^{2} - 128 \sqrt{3} x y + 64 y^{2} + 112 x - 208 \sqrt{3} y - 517 = 0

the angle of the rotation : \tan 2 α = \frac{B}{A - C} \Rightarrow

\Rightarrow α = - 30 °

x = x^{'} \cos α - y^{'} \sin α = \frac{\sqrt{3}}{2} x^{'} + \frac{1}{2} y^{'}

y = x^{'} \sin α + y^{'} \cos α = - \frac{1}{2} x^{'} + \frac{\sqrt{3}}{2} y^{'}

\Rightarrow new equation is

256 x^{2} + 160 \sqrt{3} x - 256 y - 517 = 0

y = x^{2} + \frac{5 \sqrt{3}}{8} x - \frac{517}{256}

y = {(x + \frac{5 \sqrt{3}}{16})}^{2} - \frac{37}{16}

\Rightarrow parabola y = x^{2} had been shifted \frac{5 \sqrt{3}}{16} to the

left and \frac{37}{16} down and rotated by - 30 °

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