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 ??


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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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