Question

The locus of all such points which is equidistant from (1,2) and x-axis is???
Pls answer it... It's imp

Answer 1

Answer:  The answer is $2(a-1)x-4y=a^2-4.$

Step-by-step explanation: Let P(x,y) be any point on the locus, which is equidistant from A(1,2) and B(a,0). Here, B is any point on the X-axis.

Now, distance from P to A is

$PA=\sqrt{(x-1)^2+(y-2)^2}=\sqrt{x^2-2x+1+y^2-4y+4}=\sqrt{x^2-2x+y^2-4y+5},$

and the distance from P to B is

$PB=\sqrt{(x-a)^2+(y-0)^2}=\sqrt{x^2-2ax+a^2+y^2}.$

Since P is equidistant from A and B, so we have

$PA=PB\\\\\Rightarrow \sqrt{x^2-2x+y^2-4y+5}=\sqrt{x^2-2ax+a^2+y^2+1}\\\\\Rightarrow x^2-2x+y^2-4y+5=x^2-2ax+a^2+y^2+1\\\\\Rightarrow 2(a-1)x-4y=a^2-4.$

Thus, the required locus of point 'P' is

$2(a-1)x-4y=a^2-4.$

Answer 2

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