Question

Find the equation of the locus of a point which is moving so that its distance from the axis of X is always one half its a distance from the origin.

Answer 1

Hint: First suppose the coordinate of the point. Then write the mathematical presentation of the stated condition. Use the distance formula to obtain the distance of the point and origin. Then substitute the values to obtain the required locus.

Formula Used:
The distance between the points \[(a,b)\] and \[(c,d)\] is
\[\sqrt {{{\left( {c - a} \right)}^2} + {{\left( {d - b} \right)}^2}} \] .
The distance of a point from the x axis is called the y coordinate of the point.

Complete step by step solution:
Suppose the point is \[P\left( {x,y} \right)\].
Then the given condition is
\[y = \dfrac{1}{2}OP\] --(1)
, where O is the origin.
Now,
\[OP = \sqrt {{{\left( {x - 0} \right)}^2} + {{(y - 0)}^2}} \]
\[ = \sqrt {{x^2} + {y^2}} \]
Hence, from (1) we have,
\[y = \dfrac{1}{2}\sqrt {{x^2} + {y^2}} \]
\[2y = \sqrt {{x^2} + {y^2}} \]
Square both sides of the equation,
\[4{y^2} = {x^2} + {y^2}\]
\[3{y^2} - {x^2} = 0\]

The required locus is \[3{y^2} - {x^2} = 0\].

Additional information:
Locus of a point is a set of points that satisfy an equation of curve. The curve may be a circle or hyperbola or ellipse or a line etc.
In the given question, we get a quadratic equation. It represents a pair of equations.

Note: The distance of a point from the origin is square root of the sum of squares of abscissa and ordinate of the point. By using these formulas we will calculate the distance of the point P(x,y) from the origin and equate it with the ordinate of the point P(x,y). This relation is known as the locus of the point.