Precalculus

Analytic Geometry

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Parabola

Definition and Properties of a Parabola

A parabola is a plane curve, every point of which has the property that the distance to a fixed point (called the focus of the parabola) is equal to the distance to a straight line (the directrix of the parabola). The distance from the focus to the directrix is called the focal parameter and denoted by p. A parabola has a single axis of symmetry that intersects the parabola at its vertex.

Definition of a parabola
Figure 1.

The canonical equation of a parabola has the form

\[y^2 = 2px\]

where p is the focal parameter.

Coordinates of the Vertex

\[M\left({0,0}\right)\]

Coordinates of the Focus

\[F\left({\frac{p}{2},0}\right)\]

Equation of the Directrix

\[y = -\frac{p}{2}\]

The directrix is represented by the line t in the figures.

General Equation of a Parabola

\[Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\]

where \(A,\) \(B,\) \(C,\) \(D,\) \(E,\) \(F\) are real numbers and \(B^2 - 4AC = 0.\)

Equation of a Parabola Whose Axis is Parallel to \(Y-\)axis

\[Ax^2 + Dx + Ey + F = 0\]

where \(A \ne 0,\) and \(E \ne 0,\) or in the equivalent form

\[y = ax^2 + bx + c,\;\;p=\frac{1}{2a}\]

Here \(p\) is the focal parameter.

Parabola with axis parallel to the y-axis
Figure 2.

Coordinates of the Vertex

\[M\left({x_0,y_0}\right):\;x_0 = -\frac{b}{2a},\; y_0 = ax_0^2 + bx_0 + c = \frac{4ac-b^2}{4a}\]

Coordinates of the Focus

\[F\left({x_0, y_0 + \frac{p}{2}}\right)\]

Equation of the Directrix

\[y = y_0 -\frac{p}{2}\]

Equation of a Parabola with Vertex at the Origin and Axis Parallel to \(Y-\)axis

\[y = ax^2,\;\;p=\frac{1}{2a}\]

where \(p\) is the focal parameter of the parabola.

Parabola with vertex at the origin and axis parallel to the y-axis
Figure 3.

Coordinates of the Vertex

\[M\left({0,0}\right)\]

Coordinates of the Focus

\[F\left({0,\frac{p}{2}}\right)\]

Equation of the Directrix

\[y = -\frac{p}{2}\]

Solved Problems

Example 1.

Find the coordinates of the focus and the equation of the directrix for the parabola \({y^2 = 12x.}\)

Solution.

Determine the focal parameter \(p\) of the parabola:

\[y^2 = 2px = 12x, \Rightarrow 2p = 12, \Rightarrow p = 6.\]

Calculate the coordinates of the focus \(F:\)

\[F = \left({\frac{p}{2},0}\right) = \left({\frac{6}{2},0}\right) = \left({3,0}\right).\]

The directrix equation has the form

\[y = -\frac{p}{2} = -\frac{6}{2} = -3.\]

Example 2.

Find the equation of a parabola symmetrical about the \(y-\)axis and passing through the points \(\left({0,1}\right)\) and \(\left({2,5}\right)\).

Solution.

If a parabola is symmetrical with respect to the \(y-\)axis, its general equation looks like

\[y = ax^2 + c.\]

Let's substitute the coordinates of two known points and determine the coefficients \(a\) and \(c.\)

\[\left\{ \begin{array}{l} a \cdot 0^2 + c = 1\\ a \cdot 2^2 + c = 5 \end{array} \right., \Rightarrow \left\{ \begin{array}{l} c = 1\\ 4a + c = 5 \end{array} \right., \Rightarrow \left\{ \begin{array}{l} c = 1\\ a = 1 \end{array} \right..\]

Thus the equation of the parabola has the form

\[y = x^2 + 1.\]

Example 3.

Calculate the length of the focal diameter of the parabola \(y^2 = 2px.\)

Solution.

The focal diameter of a parabola (also known as the "latus rectum") is a line segment that passes through the focus of the parabola and is perpendicular to the axis of symmetry. It is defined as the maximum chord (line segment) of the parabola. In Figure 4, it is designated as segment \(AB.\)

Focal diameter of a parabola
Figure 4.

To calculate its length, take into account that the focus F of the parabola \(y^2 = 2px\) has coordinates \(\left({\frac{p}{2},0}\right).\) The endpoints of the focal diameter A and B belong to the parabola, that is, they satisfy the equation \(y^2 = 2px.\) Let's find the y-coordinates of these points:

\[y^2 = 2px, \Rightarrow y^2 = 2p \cdot \frac{p}{2}, \Rightarrow y^2 = p^2, \Rightarrow y_{A,B} = \pm p, \Rightarrow y_A = p,\; y_B = -p.\]

Then the length of the focal diameter AB is equal to

\[AB = y_A - y_B = p - \left(-p\right) = 2p.\]

Example 4.

Find the intersection points of the parabola \(y^2 = 12x\) and ellipse \[\frac{x^2}{25} + \frac{y^2}{16} = 1.\]

Solution.

We substitute \(y^2\) from the equation of the parabola into the equation of the ellipse:

\[\frac{x^2}{25} + \frac{12x}{16} = 1, \Rightarrow \frac{x^2}{25} + \frac{3x}{4} = 1, \Rightarrow 4x^2 + 75x - 100 = 0.\]

Solve the resulting quadratic equation and find the values of x:

\[4x^2 + 75x - 100 = 0, \Rightarrow D = 75^2 - 4\cdot 4\cdot\left(-100\right) = 5625 + 1600 = 7225 = 85^2,\]
\[\Rightarrow x_{1,2} = \frac{-75 \pm \sqrt{7225}}{8} = \frac{-75 \pm 85}{8} = -20,\frac{5}{4}.\]

Note that substituting the first value \(x = -20\) into the equation of parabola gives \(y^2 = -240x,\) and we have no real solutions for y. So the real solutions are provided only by the root \(x = \frac{5}{4}.\)

\[y^2 = 12x = 12\cdot\frac{5}{4} = 15, \Rightarrow y_{1,2} = \pm\sqrt{15}.\]

So we get two intersection points of the parabola and ellipse:

\[A\left({\frac{5}{4},\sqrt{15}}\right),\;B\left({\frac{5}{4},-\sqrt{15}}\right).\]
Intersection of a parabola and an ellipse
Figure 5.