The parabola with directrix x+2y-1=0 and focu | vedclass

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(A) Let $P(x, y)$ be any point on the parabola.By the definition of a parabola,the distance from $P$ to the focus $S(1, 0)$ is equal to the perpendicu

Vedclass · Accepted Answer

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

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(A) Let $P(x, y)$ be any point on the parabola.By the definition of a parabola,the distance from $P$ to the focus $S(1, 0)$ is equal to the perpendicular distance from $P$ to the directrix $x+2y-1=0$.$PS = PM$$\sqrt{(x-1)^2 + (y-0)^2} = \frac{|x+2y-1|}{\sqrt{1^2+2^2}}$Squaring both sides,we get:$(x-1)^2 + y^2 = \frac{(x+2y-1)^2}{5}$$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 parabola with directrix $x+2y-1=0$ and focus $(1,0)$ is

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