The equation of the parabola with x+2y=1 as d | vedclass

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Question

(A) The directrix is $x+2y-1=0$ and the focus is $S(1,0)$.Let $P(x, y)$ be any point on the parabola.By definition,the distance from $P$ to the focus

Vedclass · Accepted Answer

$4x^2-4xy+y^2-8x+4y+4=0$

Vedclass · Answer

(A) The directrix is $x+2y-1=0$ and the focus is $S(1,0)$.Let $P(x, y)$ be any point on the parabola.By definition,the distance from $P$ to the focus equals the perpendicular distance from $P$ to the directrix $(PS = PM)$.$PS^2 = PM^2$$(x-1)^2 + (y-0)^2 = \frac{(x+2y-1)^2}{1^2+2^2}$$5((x-1)^2 + y^2) = (x+2y-1)^2$$5(x^2 - 2x + 1 + y^2) = x^2 + 4y^2 + 1 + 4xy - 2x - 4y$$5x^2 - 10x + 5 + 5y^2 = x^2 + 4y^2 + 4xy - 2x - 4y + 1$$4x^2 - 4xy + y^2 - 8x + 4y + 4 = 0$

The equation of the parabola with $x+2y=1$ as directrix and $(1,0)$ as focus is

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