Question

Find the equation of the locus of all points equidistant from point (2,4) and y axis.

Answer 1

Iffat Khan, Studies CSE from Rajiv Gandhi Proudyogiki Vishwavidyalaya, Bhopal (2019)

Answered Mar 1, 2016

Let the point is P(x,y)
And, it's distance from (2,4) = √[( x -2 )² + ( y - 4)²].                           ....(1)
Similarly,
It's distance from y-axis = √[( x - 0 )² + ( y - y )²]                                             
Now, according the question point P(x,y) is equidistant from given points.
So, equations (1)&(2) are equal.
Thus,
      √[( x - 2 )²+ ( y - 4 )²] = √( x )²
 On squaring both sides,
       ( x - 2 )² + ( y - 4 )² = x²
 or,   x² - 2x + 4 + y² - 8y + 16 = x²
 or,   x² - x² + y² - 2x - 8y + 20 = 0
 or,   y² - 2x - 8y + 20 = 0
Therefore the locus of the point P(x,y) is 
     y² - 2x - 8y + 20 = 0.