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Question

(B) Given point $A(a, 0)$ and the line $x+y=0$. Let point $P$ be $(h, k)$. According to the question,the distance from $P$ to $A$ equals the perpendic

Vedclass · Accepted Answer

$x^2+y^2-2xy-4ax+2a^2=0$

Vedclass · Answer

(B) Given point $A(a, 0)$ and the line $x+y=0$. Let point $P$ be $(h, k)$. According to the question,the distance from $P$ to $A$ equals the perpendicular distance from $P$ to the line $x+y=0$: $\sqrt{(h-a)^2+(k-0)^2} = \frac{|h+k|}{\sqrt{1^2+1^2}} = \frac{|h+k|}{\sqrt{2}}$. Squaring both sides,we get: $(h-a)^2 + k^2 = \frac{(h+k)^2}{2}$. $2(h^2 - 2ha + a^2 + k^2) = h^2 + k^2 + 2hk$. $2h^2 - 4ha + 2a^2 + 2k^2 = h^2 + k^2 + 2hk$. $h^2 + k^2 - 2hk - 4ha + 2a^2 = 0$. Replacing $(h, k)$ with $(x, y)$,the locus is: $x^2 + y^2 - 2xy - 4ax + 2a^2 = 0$. Thus,option $B$ is correct.

If the distance from a variable point $P$ to a fixed point $A(a, 0)$ is equal to the perpendicular distance from $P$ to the line $x+y=0$,then the equation of the locus of $P$ is

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