Question

Find the locus of the point P for which of the distance from P (4, 0) is double the distance from P to x-axis.

Answer 1

Hint: To solve the question, we have to analyse that the point P on x-axis, implies a point P with y-coordinate equal to 0. Thus, the point P on the x-axis has only x-coordinate of P. To solve further, apply the formula for distance between two points to ease the procedure of solving and to arrive at the locus of point P.

Complete step-by-step answer:
Let the point P be (x, y).
We know the formula for the distance between two points $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)$ is given by $\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$
By applying the distance between two points formula for the points P(x, y) and (4, 0)
We get

& \sqrt{{{\left( x-4 \right)}^{2}}+{{\left( y-0 \right)}^{2}}} \\\ & =\sqrt{{{\left( x-4 \right)}^{2}}+{{y}^{2}}} \\\ \end{aligned}$$ Thus, the distance between two points formula for the points P(x, y) and (4, 0) $$=\sqrt{{{\left( x-4 \right)}^{2}}+{{y}^{2}}}$$ The point on x-axis has no y-coordinate. Thus, we get the point P(x, y) on the x-axis is (x, 0). By applying the distance between two points formula for the points P(x, y) and (x, 0), we get $$\begin{aligned} & \sqrt{{{\left( x-x \right)}^{2}}+{{\left( y-0 \right)}^{2}}} \\\ & =\sqrt{{{0}^{2}}+{{y}^{2}}} \\\ & =\sqrt{{{y}^{2}}} \\\ & =y \\\ \end{aligned}$$ Thus, the distance between two point’s formula for the points P(x, y) and (x, 0) $$=y$$ Given that the distance between P(x, y) and (4, 0) is twice the distance between the point P(x, y) and x-axis, we get $$\sqrt{{{\left( x-4 \right)}^{2}}+{{y}^{2}}}=2y$$ On squaring on both sides of the equation, we get $$\begin{aligned} & {{\left( \sqrt{{{\left( x-4 \right)}^{2}}+{{y}^{2}}} \right)}^{2}}={{\left( 2y \right)}^{2}} \\\ & {{\left( x-4 \right)}^{2}}+{{y}^{2}}=4{{y}^{2}} \\\ & {{\left( x-4 \right)}^{2}}=4{{y}^{2}}-{{y}^{2}} \\\ & {{\left( x-4 \right)}^{2}}=3{{y}^{2}} \\\ \end{aligned}$$ We know the formula $${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$$ . By applying the formula to the above equation, we get $$\begin{aligned} & {{x}^{2}}+{{4}^{2}}-2\times 4\times x=3{{y}^{2}} \\\ & {{x}^{2}}+16-8x=3{{y}^{2}} \\\ & {{x}^{2}}-3{{y}^{2}}-8x+16=0 \\\ \end{aligned}$$ Thus, the locus of point P(x, y) is $${{x}^{2}}-3{{y}^{2}}-8x+16=0$$ Note: The possibility of mistake can be, not applying the formula for distance between two points, which is important to ease the procedure of solving. The other possibility of mistake can be, not being able to analyse the point P on the x-axis, which implies a point P with y-coordinate equal to 0.