Question

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

Answer 1

Step-by-step explanation:

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 - 4x + 4 + y² - 8y + 16 = x²

Because,

( a - b )² = a² - 2ab + b²

or, x² - x² + y² - 4x - 8y + 20 = 0

or, y² - 4x - 8y + 20 = 0

Therefore the locus of the point P(x,y) is

y² - 4x - 8y + 20 = 0.