Conics

Lessons

Senior 6 · Mathematics

Conics

5 lessons

Lesson 2 of 5

The parabola

40 min

The curve with one focus

The dish in the photograph below collects radio waves from a satellite tens of thousands of kilometres away. The waves arrive almost parallel, and the dish must send all of them to one small receiver. The only curve that does this is the parabola: every ray parallel to its axis is reflected through the same point, the focus. This lesson finds its equation, its tangents and normals, and proves that reflecting property.

Definition

Parabola: the set of points $P$ equidistant from a fixed point $F$ , the focus, and a fixed line $d$ , the directrix, not through $F$ . It is the conic with eccentricity $e = 1$ .

The line through $F$ perpendicular to $d$ is the axis; the point where the axis meets the parabola, midway between $F$ and $d$ , is the vertex. The chord through $F$ perpendicular to the axis is the latus rectum. Example: $y^{2} = 8 x$ has focus $(2, 0)$ , directrix $x = - 2$ and vertex $(0, 0)$ .

Important

Standard equation. Put the vertex at the origin, the focus at $F (a, 0)$ with $a \gt 0$ , and the directrix $x = - a$ . For $P (x, y)$ , $P F = P M$ reads

\sqrt{(x - a)^2 + y^2} = |x + a| \iff x^2 - 2ax + a^2 + y^2 = x^2 + 2ax + a^2 \iff y^2 = 4ax .

Its latus rectum joins $(a, - 2 a)$ to $(a, 2 a)$ , so it has length $4 a$ . The focal distance of a point on the curve is $P F = P M = x + a$ . A convenient parametric form is $x = a t^{2}$ , $y = 2 a t$ , $t \in \mathbb{R}$ .

Equation	Opens	Focus	Directrix	Axis
$y^{2} = 4 a x$	right	$(a, 0)$	$x = - a$	$y = 0$
$y^{2} = - 4 a x$	left	$(- a, 0)$	$x = a$	$y = 0$
$x^{2} = 4 a y$	up	$(0, a)$	$y = - a$	$x = 0$
$x^{2} = - 4 a y$	down	$(0, - a)$	$y = a$	$x = 0$
$(y - k)^{2} = 4 a (x - h)$	right	$(h + a, k)$	$x = h - a$	$y = k$

In every row $a \gt 0$ is the distance from the vertex to the focus. The last row is $y^{2} = 4 a x$ moved so that its vertex is at $(h, k)$ .

Worked example 1: reading the elements

Give the focus, directrix and length of the latus rectum of (a) $y^{2} = - 12 x$ (b) $x^{2} = 6 y$ .

Solution. (a) $4 a = 12$ , $a = 3$ , and the minus sign means it opens to the left: focus $(- 3, 0)$ , directrix $x = 3$ , latus rectum 12. (b) $4 a = 6$ , $a = \frac{3}{2}$ , opening upwards: focus $\left(0, \frac{3}{2}\right)$ , directrix $y = -\frac{3}{2}$ , latus rectum 6.

Worked example 2: the equation from the focus and directrix

Find the parabola with focus $(3, - 1)$ and directrix $x = - 1$ .

Solution. The axis is $y = - 1$ . The vertex is midway between the focus and the directrix, at $(1, - 1)$ , and $a = 3 - 1 = 2$ ; the focus is to the right of the directrix, so it opens to the right: $(y + 1)^{2} = 8 (x - 1)$ .

Check with the definition: $(x - 3)^{2} + (y + 1)^{2} = (x + 1)^{2}$ gives $(y + 1)^{2} = (x + 1)^{2} - (x - 3)^{2} = 8 x - 8$ .

Tangent and normal

Differentiate $y^{2} = 4 a x$ implicitly: $2y\,\frac{dy}{dx} = 4a$ , so at $P (x_{1}, y_{1})$ with $y_1 \ne 0$ the slope of the tangent is $\frac{2a}{y_1}$ . The tangent is $y - y_1 = \frac{2a}{y_1}(x - x_1)$ ; multiply by $y_{1}$ and use $y_{1}^{2} = 4 a x_{1}$ :

\text{tangent at } (x_1, y_1):\quad yy_1 = 2a(x + x_1) , \qquad \text{normal:}\quad y - y_1 = -\frac{y_1}{2a}(x - x_1) .

(At the vertex the tangent is the $y$ -axis.) In parametric form, at $(a t^{2}, 2 a t)$ : the tangent is $t y = x + a t^{2}$ and the normal is $y + t x = 2 a t + a t^{3}$ .

Tangent of given slope. Substituting $y = m x + k$ into $y^{2} = 4 a x$ gives a quadratic in $x$ ; the line is a tangent when it has a double root, which happens exactly when $k = \frac{a}{m}$ ( $m \ne 0$ ). The point of contact is $\left(\frac{a}{m^2}, \frac{2a}{m}\right)$ .

The parabola

y^{2} = 8 x

, drawn from its equation, with its focus, directrix, a tangent, a normal and one reflected ray.

Worked example 3: tangent and normal at a point

Find the tangent and the normal to $y^{2} = 8 x$ at $P (2, 4)$ . Where does the normal meet the parabola again?

Solution. Here $a = 2$ . Tangent: $4 y = 4 (x + 2)$ , that is $y = x + 2$ . Its slope is 1, so the normal has slope $- 1$ : $y - 4 = - (x - 2)$ , that is $y = 6 - x$ .

Substitute into $y^{2} = 8 x$ : $(6 - x)^2 = 8x \iff x^2 - 20x + 36 = 0 \iff (x - 2)(x - 18) = 0$ . Besides $P$ , the normal meets the curve at $(18, - 12)$ . Check: $(-12)^2 = 144 = 8 \times 18$ .

Worked example 4: a tangent with a given slope

Find $k$ so that $y = 2 x + k$ is a tangent to $y^{2} = 8 x$ , and the point of contact.

Solution. Substitute: $(2x + k)^2 = 8x \iff 4x^2 + (4k - 8)x + k^2 = 0$ . A tangent needs a zero discriminant: $(4k - 8)^2 - 16k^2 = 0 \iff -64k + 64 = 0 \iff k = 1$ , which agrees with $k = \frac{a}{m} = \frac{2}{2}$ . Then $4 x^{2} - 4 x + 1 = 0$ , $x = \frac{1}{2}$ , $y = 2$ : the point of contact is $\left(\frac{1}{2}, 2\right)$ .

Important

Reflecting property. The tangent at any point $P$ of a parabola makes equal angles with $P F$ and with the line through $P$ parallel to the axis. So a ray travelling parallel to the axis is reflected through the focus, and a lamp at the focus sends out a parallel beam.

Proof. The tangent $y y_{1} = 2 a (x + x_{1})$ meets the axis $y = 0$ at $T (- x_{1}, 0)$ . Then $F T = x_{1} + a$ , and $F P = x_{1} + a$ (the focal distance). Triangle $F T P$ is isosceles, so $\angle FTP = \angle FPT$ . The line through $P$ parallel to the axis makes with the tangent an angle equal to $\angle FTP$ (corresponding angles). In the figure, $P (4.5, 6)$ , $T (- 4.5, 0)$ and $F T = F P = 6.5$ .

Worked example 5: where to put the receiver

A dish is $2.4\,\mathrm{m}$ across its rim and $0.3\,\mathrm{m}$ deep at the centre. How far from the vertex must the receiver be placed?

Solution. Model a cross-section by $x^{2} = 4 a y$ with the vertex at the origin. The rim is at $(1.2, 0.3)$ : ${1.2}^{2} = 4 a (0.3)$ , so $a = \frac{1.44}{1.2} = 1.2$ . The receiver goes at the focus, $1.2\,\mathrm{m}$ from the vertex along the axis.

Warning

$a$ is not the coefficient. In $y^{2} = 8 x$ , $4 a = 8$ , so $a = 2$ , not 8. And in $y = 2 x^{2}$ , first rewrite as $x^2 = \frac{1}{2}y$ : $4a = \frac{1}{2}$ , $a = \frac{1}{8}$ , so the focus is $\left(0, \frac{1}{8}\right)$ .

The tangent formula needs the point to be ON the curve. $y y_{1} = 2 a (x + x_{1})$ used with a point off the parabola gives a different line (the polar of the point), not a tangent.

Tip

The tangent formulas of all the conics follow one rule: in the equation, replace $y^{2}$ by $y y_{1}$ , $x^{2}$ by $x x_{1}$ , $x$ by $\frac{x + x_1}{2}$ and $y$ by $\frac{y + y_1}{2}$ . For $y^{2} = 4 a x$ this gives $yy_1 = 4a\cdot\frac{x + x_1}{2}$ , the formula above.

Khan Academy: Equation for parabola from focus and directrix (about 8 min). Watch the distance to the focus being set equal to the distance to the directrix and squared, as in the derivation of y² = 4ax in this lesson.

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Practice

Questions 1 to 3 are routine; 4 to 6 need more care; 7 asks for a proof.

Give the vertex, focus and directrix of (a) $y^{2} = 20 x$ (b) $x^{2} = - 8 y$ (c) $(x - 1)^{2} = 12 (y + 2)$ .
Find the equation of the parabola with focus $(0, - 3)$ and directrix $y = 3$ .
Find the tangent and the normal to $y^{2} = 12 x$ at $(3, 6)$ .
Find the point of $y^{2} = 12 x$ at which the tangent is parallel to $y = 3 x - 1$ , and the equation of that tangent.
Show that the line $y = x + 1$ is a tangent to $y^{2} = 4 x$ , and find the point of contact.
A car headlight has a parabolic reflector 16 cm across and 4 cm deep. How far from the vertex should the bulb be?
Prove that the tangents at the two ends of the latus rectum of $y^{2} = 4 a x$ meet on the directrix, at right angles.

Show answersHide answers

(a) vertex $(0, 0)$ , focus $(5, 0)$ , directrix $x = - 5$ (b) vertex $(0, 0)$ , focus $(0, - 2)$ , directrix $y = 2$ (c) vertex $(1, - 2)$ , focus $(1, 1)$ , directrix $y = - 5$
$x^{2} = - 12 y$
Tangent $6 y = 6 (x + 3)$ , so $y = x + 3$ ; normal $y = - x + 9$ .
$m = 3$ , $a = 3$ : $k = \frac{a}{m} = 1$ ; tangent $y = 3 x + 1$ , touching at $\left(\frac{1}{3}, 2\right)$ .
$(x + 1)^2 = 4x \iff (x - 1)^2 = 0$ : a double root, so a tangent, touching at $(1, 2)$ .
$x^{2} = 4 a y$ through $(8, 4)$ : $64 = 16 a$ , $a = 4$ : 4 cm from the vertex, which here is at the level of the rim.
At $(a, \pm 2a)$ the tangents are $y = x + a$ and $y = - x - a$ , with slopes $1$ and $- 1$ (product $- 1$ ); they meet at $(- a, 0)$ , on $x = - a$ .

Sources and credits

Written by Ganzaa: the explanations, the proofs, the diagram (plotted from $y^{2} = 8 x$ and the computed tangent and normal), the worked examples, the practice items and the quiz questions.
Photo: Erdfunkstelle Raisting 2, Richard Bartz, Wikimedia Commons, CC BY-SA 2.5.
Photo: Parabolic Solar Cooker, Solarbroman, Wikimedia Commons, CC BY-SA 4.0.
Video: Equation for parabola from focus and directrix, Khan Academy, embedded from YouTube.
This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1The curve with one focus

2Tangent and normal

The curve with one focus

Tangent and normal