Question

The equation to the locus of a point P for which the distance from P to (0,5) is double the distance
from P to y-axis is
1) 3r + y + 10y - 25 = 0
2) 3r - y2 + 10y + 25 = 0
3) 32- + 10y - 25 = 0
4) 3 x + y² - 10y - 25 = 0​

Answer 1

let the point be P(x,y)..

now

x^2+(y-5)^2={2(y-5)}^2

x^2+y^2+25-10y=(2y-10)^2

x^2+y^2+25-10y=4y^2+100-40y

x^2-3y^2+30y-75=0

hope it helps you dude

Answer 2

3) 3x²-y² + 10y - 25 = 0

Explanation:

Let P(x,y) be any point on x-y plane.

The distance of P from A(0,5) can be calculated as

PA=$\sqrt{(x-0)^2+(y-5)^2}$

As per the information given in this question, the distance (PA) is double the distance of P from the y-axis ( it can be assumed as $d_y$)

This means,

$PA=2d_y$

$d_y$ can be taken as |x|. (∵ distance can never be negative, so the mod is there)

So,

$\sqrt{(x-0)^2+(y-5)^2}=2|x|$

On squaring both sides, we get,

$x^2+(y-5)^2=4x^2\\\\\implies x^2+y^2+25-10y=4x^2\\\\\implies -3x^2+y^2-10y+25=0\\\\\implies 3x^2-y^2+10y-25=0$

Hence the required equation to the locus of a point P will be  3x² - y² + 10y - 25 = 0​.

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