The equation of the parabola with focus (0,0) | vedclass

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Question

(A) The focus of the parabola is $S(0,0)$.The equation of the directrix is $x+y-4=0$.Let $P(x, y)$ be any point on the parabola.By definition,the dist

Vedclass · Accepted Answer

$x^2+y^2-2xy+8x+8y-16=0$

Vedclass · Answer

(A) The focus of the parabola is $S(0,0)$.The equation of the directrix is $x+y-4=0$.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,so $SP^2 = PM^2$.$(x-0)^2 + (y-0)^2 = \left(\frac{x+y-4}{\sqrt{1^2+1^2}}\right)^2$$x^2 + y^2 = \frac{(x+y-4)^2}{2}$$2(x^2 + y^2) = (x+y-4)^2$$2x^2 + 2y^2 = x^2 + y^2 + 16 + 2xy - 8x - 8y$$x^2 + y^2 - 2xy + 8x + 8y - 16 = 0$

The equation of the parabola with focus $(0,0)$ and directrix $x+y=4$ is

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