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

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Question

(A) The definition of a parabola is the locus of a point $P(x, y)$ such that its distance from the focus is equal to its perpendicular distance from t

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The definition of a parabola is the locus of a point $P(x, y)$ such that its distance from the focus is equal to its perpendicular distance from the directrix.Given focus $S(1, 0)$ and directrix $x+2y-1=0$.Using the distance formula,the equation is:$\frac{|x+2y-1|}{\sqrt{1^2+2^2}} = \sqrt{(x-1)^2 + (y-0)^2}$Squaring both sides:$\frac{(x+2y-1)^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$Rearranging the terms to one side:$4x^2 - 4xy + y^2 - 8x + 4y + 4 = 0$Thus,option $A$ is correct.

The parabola with directrix $x+2y-1=0$ and focus $(1,0)$ is

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