Question

Find the locus of the point whose distance from x axis equals 2/3 times its distance from the origin.

Answer 1

SOLUTION

Let the point be P= P(x,y).

Then, distance of P(x,y) from x-axis.

=) 2/3 × distance of P(x,y) from the origin

=) y= 2/3√(x-0)^2 + (y-0)^2

=) 2/3√x^2+ y^2

On squaring both the sides, we get

$= > {y}^{2} = \frac{4}{9}( {x}^{2} + {y}^{2}) \\ = > 9 {y}^{2} = 4( {x}^{2} + {y}^{2} ) \\ = > 9 {y}^{2} - 4x {}^{2} - 4 {y}^{2} = 0 \\ = > 4 {x}^{2} - 5 {y}^{2} = 0$

which is the required locus of the point.

hope it helps ✔️

Answer 2

Step-by-step explanation:

hope it helps you !...................