Precalculus

Analytic Geometry

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Equation of a Plane Curve

Equation of a Plane Curve in Cartesian Coordinates

Let a Cartesian coordinate system and some curve C be given on a plane. Consider an equation relating two variables x and y:

\[f\left({x,y}\right) = 0\]

The above equation is called the implicit equation of the curve C, if the coordinates (x, y) of any point lying on the curve satisfy this equation and, respectively, the coordinates (x, y) of any point not lying on the curve do not satisfy this equation.

So for example the equation of a circle of radius R centered at the point M(a, b) has the form

\[\left({x - a}\right)^2 + \left({y - b}\right)^2 = R^2\]
A circle of radius R centered at the point (a,b)
Figure 1.

If an equation of a curve can be solved for one of the variables y or x, then we get an explicit form of the equation:

\[y = f\left({x}\right) \text{ or } x = g\left({y}\right)\]

Parametric Representation of a Curve

It is often convenient to express the coordinates of the points (x ,y) of a curve C using the third auxiliary variable t (called a parameter):

\[x = \varphi\left({t}\right),\;\;y = \psi\left({t}\right)\]

where the functions φ(t) and ψ(t) are assumed to be continuous in some range of the parameter t.

Note that by eliminating the parameter t from both these equations we can find a relationship between x and y, that is the general equation of the curve in the form

\[f\left({x,y}\right) = 0.\]

If the variable t is the time counted from some initial moment, then the functions φ(t) and ψ(t) determine the law of motion of the body.

As an example, we write the parametric equations of an ellipse:

\[x = a\cos t,\;\;y = b\sin t\]
An ellipse with semi-axes a and b
Figure 2.

The ellipse equation in parametric form can be easily converted to standard form:

\[\left\{\begin{array}{*{20}{l}} x = a\cos t\\ y = b\sin t \end{array} \right.,\;\Rightarrow \left\{\begin{array}{*{20}{l}} \frac{x}{a} = \cos t\\ \frac{y}{b} = \sin t \end{array} \right.,\;\Rightarrow \left\{\begin{array}{*{20}{l}} \frac{x^2}{a^2} = \cos^2 t\\ \frac{y^2}{b^2} = \sin^2 t \end{array} \right.,\;\Rightarrow \frac{x^2}{a^2} + \frac{y^2}{b^2} = \cos^2 t + \sin^2 t = 1.\]

Equation of a Curve in Polar Coordinates

The form of the equation of a curve C depends not only on the curve itself, but also on the choice of the coordinate system.

If a curve in the Cartesian coordinate system is described by the equation f(x, y) = 0, then in order to obtain the equation of the same curve with respect to any other coordinate system, it is enough to substitute instead of x and y their expressions in terms of the new coordinates. In case of polar coordinates, the variables x and y are expressed in terms of polar coordinates ρ and φ as follows:

\[x = \rho\cos\varphi,\;\;y = \rho\sin\varphi\]

Substituting these relations into the equation of the curve in Cartesian coordinates, we obtain its equation in polar coordinates in the form

\[f\left({\rho\cos\varphi,\rho\sin\varphi}\right) = 0,\;\Rightarrow \Phi\left({\rho,\varphi}\right) = 0\]

Many curves have a simpler equation in polar coordinates than in Cartesian coordinates. Let's take a cardioid as an example.

Cardioid
Figure 3.

The cardioid shown in the figure is described by a simple polar equation

\[\rho\left({\varphi}\right) = 1 + \cos\varphi\]

The equation describing this curve in Cartesian coordinates has a more complex form:

\[\left({x^2 + y^2}\right)^2 - 2x\left({x^2 + y^2}\right) - y^2 = 0.\]

Solved Problems

Click or tap a problem to see the solution.

Example 1

A point M moves along a curve so that it is equidistant from points A(3,0) and B(-3,6). Find the equation of the curve.

Example 2

A point M moves along a curve so that it is equidistant from the origin and point B(-4,2). Write the equation of the curve.

Example 3

Find the locus of points equidistant from the y-axis and point B(2,0).

Example 4

Points A and B are given on a plane. Find the locus of points M that are twice as distant from A than from B.

Example 1.

A point M moves along a curve so that it is equidistant from points A(3,0) and B(-3,6). Find the equation of the curve.

Solution.

Let the coordinates of the point M be (x,y). Since

\[d\left(M,A\right) = d\left(M,B\right),\]

we get

\[\sqrt{\left({x - 3}\right)^2 + \left({y - 0}\right)^2} = \sqrt{\left({x + 3}\right)^2 + \left({y - 6}\right)^2}.\]

Simplify this equation:

\[\left({x - 3}\right)^2 + y^2 = \left({x + 3}\right)^2 + \left({y + 6}\right)^2, \Rightarrow \cancel{x^2} - 6x + \cancel{9} + \cancel{y^2} = \cancel{x^2} + 6x + \cancel{9} + \cancel{y^2} - 12y + 36,\]
\[\Rightarrow 12y = 12x + 36, \Rightarrow y = x + 3.\]

We got the equation of a straight line.

Example 2.

A point M moves along a curve so that it is equidistant from the origin and point B(-4,2). Write the equation of the curve.

Solution.

Locus of points that are equidistant from the origin and point (-4,2)
Figure 4.

Assuming that M(x,y) is an arbitrary point of the curve, we can write:

\[d\left(M,A\right) = d\left(M,B\right).\]

Substituting the known coordinates, we get

\[\sqrt{\left({x + 4}\right)^2 + \left({y - 2}\right)^2} = \sqrt{x^2 + y^2}, \Rightarrow \cancel{x^2} + 8x + 16 + \cancel{y^2} - 4y + 4 = \cancel{x^2} + \cancel{y^2},\]

or

\[4y = 8x + 20, \Rightarrow y = 2x + 5.\]

It can be seen that this curve is a straight line.

Example 3.

Find the locus of points equidistant from the y-axis and point B(2,0).

Solution.

Let M(x,y) be an arbitrary point of the given set.

Locus of points that are equidistant from the y-axis and point (2,0)
Figure 5.

By the condition of the problem,

\[d\left(M,A\right) = d\left(M,B\right).\]

Therefore

\[\sqrt{\left({x - 0}\right)^2 + \left({y - y}\right)^2} = \sqrt{\left({x - 2}\right)^2 + \left({y - 0}\right)^2}, \Rightarrow x^2 = \left({x - 2}\right)^2 + y^2.\]

Solve the equation for x:

\[\Rightarrow \cancel{x^2} = \cancel{x^2} - 4x + 4 + y^2, \Rightarrow 4x = y^2 + 4, \Rightarrow x = \frac{y^2}{4} + 1.\]

We obtained a parabola equation. Its graph is rotated \(90^\circ\) clockwise from the standard position and opens to the right.

Example 4.

Points A and B are given on a plane. Find the locus of points M that are twice as distant from A than from B.

Solution.

Let the origin be at point A and the x-axis pass through point B. Suppose point B has coordinates (b,0).

Locus of points that are twice as far away from one point than from another
Figure 6.

It follows from the condition that

\[d\left(M,A\right) = 2d\left(M,B\right).\]

Using the distance formula between two points, we have

\[\sqrt{\left({x - 0}\right)^2 + \left({y - 0}\right)^2} = 2\sqrt{\left({x - b}\right)^2 + \left({y - 0}\right)^2}.\]

Simplify this expression:

\[x^2 + y^2 = 4\left[{\left({x - b}\right)^2 + y^2}\right], \Rightarrow x^2 + y^2 = 4\left({x^2 - 2bx + b^2}\right) + 4y^2, \Rightarrow x^2 + y^2 = 4x^2 - 8bx + 4b^2 + 4y^2,\]
\[\Rightarrow 3x^2 - 8bx + 3y^2 + 4b^2 = 0, \Rightarrow x^2 - \frac{8b}{3}x + y^2 + \frac{4b^2}{3} = 0\]

Complete the square for the terms with x:

\[x^2 - \frac{8b}{3}x + \underbrace{\left({\frac{4b}{3}}\right)^2 - \left({\frac{4b}{3}}\right)^2}_{0} + y^2 + \frac{4b^2}{3} = 0, \Rightarrow \left({x - \frac{4b}{3}}\right)^2 + y^2 = \frac{16b^2}{9} - \frac{4b^2}{3},\]

or

\[\left({x - \frac{4b}{3}}\right)^2 + y^2 = \left({\frac{2b}{3}}\right)^2.\]

We see that the locus of points is a circle of radius \({\frac{2b}{3}}\) centered at \(\left({\frac{4b}{3}, 0}\right).\)