Question

Find the locus of the point, the absolute value of the difference of the distances of
which from the points (2,2) and (0,0) is 2 . Identify the curve represented by the
locus.

Answer 1

Answer on Question #78659 – Math – Analytic Geometry

Question

Find the locus of the point, the absolute value of the difference of the distances of which from the points $(2,2)$ and $(0,0)$ is 2. Identify the curve represented by the locus.

Solution

Choose some point having coordinates $(x,y)$.

The distance between this point and $(2,2)$ is given by

$$\sqrt{(x - 2)^2 + (y - 2)^2}$$

The distance between point $(x,y)$ and $(0,0)$ is given by

$$\sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{x^2 + y^2}$$

The absolute value of the difference of the distances is 2

$$\left| \sqrt{(x - 2)^2 + (y - 2)^2} - \sqrt{x^2 + y^2} \right| = 2$$ $$\sqrt{(x - 2)^2 + (y - 2)^2} - \sqrt{x^2 + y^2} = 2$$ $$\sqrt{(x - 2)^2 + (y - 2)^2} = 2 + \sqrt{x^2 + y^2}$$ $$x^2 - 4x + 4 + y^2 - 4y + 4 = 4 + 4\sqrt{x^2 + y^2} + x^2 + y^2$$ $$\sqrt{x^2 + y^2} = -x - y + 1$$ $$x^2 + y^2 = x^2 + y^2 + 2xy - 2x - 2y + 1, -x - y + 1 \geq 0$$ $$2y(x - 1) = 2x - 1$$ $$y = \frac{2x - 1}{2(x - 1)}, x \neq 1$$ $$\sqrt{(x - 2)^2 + (y - 2)^2} - \sqrt{x^2 + y^2} = -2$$ $$\sqrt{(x - 2)^2 + (y - 2)^2} = -2 + \sqrt{x^2 + y^2}$$ $$x^2 - 4x + 4 + y^2 - 4y + 4 = 4 - 4\sqrt{x^2 + y^2} + x^2 + y^2, x^2 + y^2 \geq 2$$ $$\sqrt{x^2 + y^2} = x + y - 1$$ $$x^2 + y^2 = x^2 + y^2 + 2xy - 2x - 2y + 1, x + y - 1 \geq 0$$ $$2y(x - 1) = 2x - 1$$ $$y = \frac{2x - 1}{2(x - 1)}, x \neq 1$$

A hyperbola is the locus of a point which moves in the plane in such a way that the absolute value of the difference of its distances from two fixed points in the plane is constant.

Graph of the locus:

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