The equation of the parabola with focus (1, - | vedclass

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Question

(A) Let $P(x, y)$ be any point on the parabola. The distance of $P$ from the focus $S(1, -1)$ is equal to its perpendicular distance from the directri

Vedclass · Accepted Answer

$x^2+y^2-10x-2y-2xy-5=0$

Vedclass · Answer

(A) Let $P(x, y)$ be any point on the parabola. The distance of $P$ from the focus $S(1, -1)$ is equal to its perpendicular distance from the directrix $x+y+3=0$. $	herefore PS = PQ \implies PS^2 = PQ^2$ Using the distance formula and the perpendicular distance formula: $(x-1)^2 + (y+1)^2 = \left(\frac{x+y+3}{\sqrt{1^2+1^2}}ight)^2$ $x^2 - 2x + 1 + y^2 + 2y + 1 = \frac{(x+y+3)^2}{2}$ $2(x^2 + y^2 - 2x + 2y + 2) = x^2 + y^2 + 9 + 2xy + 6y + 6x$ $2x^2 + 2y^2 - 4x + 4y + 4 = x^2 + y^2 + 2xy + 6x + 6y + 9$ $x^2 + y^2 - 2xy - 10x - 2y - 5 = 0$

The equation of the parabola with focus $(1, -1)$ and directrix $x+y+3=0$ is

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