11.1 Parabolas: focus, directrix, tangents, and normals

Motivation

A parabola can be specified by distances before it is described by an equation. This geometric definition explains why a particular quadratic appears and why translations or changes of orientation preserve the same kind of curve. Its parameter then turns tangent and normal problems into equations in one real variable, where the number and distinctness of roots have direct geometric meaning.

The main proof obligations are easy to overlook: squaring a distance equation requires a converse, a slope formula can miss a vertical tangent, and a locus calculation must show that every claimed point is actually attained. We will keep these questions explicit rather than treating an eliminated equation as a complete geometric proof.

Focus, directrix, and the standard equation

Definition

Focus-directrix description of a parabola

Let $F$ be a fixed point not on a fixed genuine line $D$ in the plane. The focus-directrix condition with eccentricity $e\gt0$ is $\operatorname{dist}(X,F)=e\operatorname{dist}(X,D)$. The point $F$ is the focus and $D$ is the directrix. A parabola is the case $e=1$, so the two distances are equal.

The distance to a line is perpendicular distance, not distance to an arbitrary point on it. Since $F\notin D$, a point satisfying the condition cannot lie on $D$: that would force its distance to $F$ to be zero as well. Thus the distance ratio is defined on the locus. A genuine line has a nonzero normal; an empty set or the whole plane is not a directrix.

Choose the horizontal axis through $F$ perpendicular to $D$, with origin midway between $F$ and its perpendicular foot on $D$. Then we can write $F=(a,0)$ and $D:x=-a$, where $a\gt0$. The separation of focus and directrix is $2a$; positivity records their distinctness.

Concept lensGeometric

The equation records an equality of distances

For $X=(x,y)$, distance to the focus is $\sqrt{(x-a)^2+y^2}$, while distance to the vertical directrix is $|x+a|$. The first quantity measures a point-to-point displacement; the second measures only the horizontal component needed to reach a line. The equation emerges by equating these two different geometric measurements.

The origin lies on the locus because its two distances are both $a$. The equation below will show that every other point has $x\ge0$, making the origin the vertex and the horizontal axis the axis of symmetry.

Theorem

The focus-directrix equation is equivalent to the parabola equation

For $a\gt0$, $F=(a,0)$, and $D:x=-a$, a point $X=(x,y)$ has equal distances to $F$ and $D$ if and only if

$$y^2=4ax.$$

Equality of the distances gives

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

Conversely, if $y^2=4ax$, then $x\ge0$ because $a\gt0$. Substitution gives $(x-a)^2+y^2=(x+a)^2$. Taking nonnegative square roots yields exactly $\sqrt{(x-a)^2+y^2}=|x+a|$; in this case $x+a\gt0$ as well. Thus no extraneous points have been introduced by squaring. The symmetry under $y\mapsto-y$ and the unique point with $x=0$ are now visible algebraically.

A translation gives $(y-k)^2=4a(x-h)$, with vertex $(h,k)$, focus $(h+a,k)$, and directrix $x=h-a$. Replacing the right side by $-4a(x-h)$ reverses the opening: its focus is $(h-a,k)$ and directrix $x=h+a$. Interchanging the coordinate roles gives the upward form $(x-h)^2=4a(y-k)$, with focus $(h,k+a)$ and directrix $y=k-a$; the downward form has a minus sign, focus $(h,k-a)$, and directrix $y=k+a$. Here $a$ remains positive; the sign in the equation specifies orientation.

t = 1

P = (1, 2); PF = PD = 2

For $y^2=4x$, the focus is $F=(1,0)$ and the directrix is $x=-1$. The point $P=(t^2,2t)$ has $PF=PD=t^2+1$, where $D$ is its perpendicular foot on the directrix. The fine dashed line is the tangent $x-ty+t^2=0$. At $t=0$ it is vertical: $x=0$.

Parameters, tangents, and normals

Every point of the standard parabola can be written uniquely as

$$P(t)=(at^2,2at),\qquad t\in\mathbb R.$$

Substitution verifies membership. Conversely, for a point $(x,y)$ on the curve, $t=y/(2a)$ gives $at^2=y^2/(4a)=x$. Thus the parameter neither omits the vertex nor represents a point more than once.

Theorem

Tangent and normal equations valid at every real parameter

At $P(t)=(at^2,2at)$, the tangent and normal are respectively

$$x-ty+at^2=0,\qquad tx+y-2at-at^3=0.$$

These equations are valid for every $t\in\mathbb R$, including $t=0$.

Differentiate the parametrization to get $P'(t)=(2at,2a)$. It never vanishes because $a\gt0$, so $(t,1)$ is a tangent direction and $(1,-t)$ is a perpendicular normal direction. The tangent is therefore $(1,-t)\cdot((x,y)-P(t))=0$, which gives its equation. The normal line through the same point has normal vector $(t,1)$; expanding its point-normal equation gives the second formula.

For $t\ne0$, the tangent slope is $1/t$ and the normal slope is $-t$. At $t=0$, the tangent is $x=0$ and the normal is $y=0$. The line equations remain meaningful where the slope calculation cannot be used. For a contact point $(x_0,y_0)$, equivalent forms are

$$y_0y=2a(x+x_0),\qquad y_0(x-x_0)+2a(y-y_0)=0.$$

These follow by substituting $x_0=at^2,y_0=2at$; they too include the vertex.

Worked example

Translate the geometry and retain the vertex tangent

For $(y+2)^2=8(x-3)$, we have $a=2$ and vertex $(3,-2)$. The focus is $(5,-2)$ and the directrix is $x=1$. A parametrization is $(3+2t^2,-2+4t)$, whose derivative is $(4t,4)$. At $t=0$ the tangent direction is vertical, so the tangent is $x=3$; the perpendicular normal is $y=-2$. No finite tangent slope exists there, but neither the curve nor either line is undefined.

Tangents from a point and the chord of contact

A tangent at parameter $t$ passes through $H=(h,k)$ exactly when

$$at^2-kt+h=0.$$

Because $a\gt0$, this is a genuine quadratic. Its discriminant $\Delta=k^2-4ah$ counts the real contact parameters: two distinct roots give two tangents, a repeated root gives one tangent, and no real roots give no real tangent. Distinct parameters give distinct lines because the normalized tangent equation has coefficient $-t$ on $y$. When $\Delta=0$, the point itself satisfies $k^2=4ah$ and lies on the parabola.

Suppose $\Delta\gt0$, so $H$ is external in the sense of admitting two distinct tangents. If their contact points are $(x_1,y_1),(x_2,y_2)$, substituting $H$ into each coordinate tangent equation yields $2a(h+x_i)-ky_i=0$. Consequently their chord of contact is

$$2a(x+h)-ky=0.$$

This is a genuine line because its $x$ coefficient is $2a\ne0$. It contains two distinct contact points, which determine it uniquely. No division by $k$ is needed; when $k=0$, it is the vertical line $x=-h$. Without two distinct tangents, the geometric chord of contact has not been established even though the displayed linear expression can still be written.

Worked example

Two tangents through an external point

For $y^2=4x$ and $H=(-2,1)$, the parameter equation is $t^2-t-2=(t-2)(t+1)=0$. The contact parameters are $2,-1$, giving

$$x-2y+4=0,\qquad x+y+1=0.$$

Their contact points are $(4,4)$ and $(1,-2)$. Both tangents pass through $(-2,1)$, and the chord of contact is $2x-y-4=0$. Substitution of each contact point verifies this chord; it is a different line from either tangent, since it joins the two points of tangency.

Reflection toward the focus

Consider light travelling parallel to the axis toward the parabola with unit incoming direction $d=(-1,0)$. At $P(t)$, reflection in the tangent preserves the tangential component of this direction and reverses its normal component. This vector description avoids special cases where a tangent or reflected ray is vertical.

Proof: Reflect the normal component without dividing by a slope

Normal and legal denominator. A normal to the tangent is $n=(1,-t)$, so $\|n\|^2=1+t^2\gt0$ for every real $t$. The normal component of $d$ is $((d\cdot n)/\|n\|^2)n$.

Reflection step. Reversing that component subtracts it twice:

$$d_{\mathrm{out}} =d-2\frac{d\cdot n}{\|n\|^2}n =\frac{(1-t^2,-2t)}{1+t^2}.$$

Target direction. With $F=(a,0)$,

$$F-P(t)=a(1-t^2,-2t),\qquad \|F-P(t)\|=a(1+t^2).$$

Thus $d_{\mathrm{out}}=(F-P(t))/\|F-P(t)\|$: the reflected ray travels toward the focus, rather than merely along an unoriented line through it.

Boundary check. At $t=0$ the outgoing direction is $(1,0)$; at $t=1$ it is $(0,-1)$; at $t=-1$ it is $(0,1)$. The same calculation covers all three cases without the forbidden divisions by $t$ or $t^2-1$ that a slope argument might introduce.

A midpoint locus from perpendicular radius vectors

Let $O=(0,0)$ and let distinct contacts $A=P(t_1)$ and $B=P(t_2)$ satisfy $\angle AOB=\pi/2$. Both contacts must differ from $O$, so $t_1t_2\ne0$. Set $s=t_1+t_2$ and $p=t_1t_2$. Dot products give

$$\overrightarrow{OA}\cdot\overrightarrow{OB} =a^2t_1t_2(t_1t_2+4)=a^2p(p+4)=0.$$

Since $p\ne0$, necessarily $p=-4$. Merely accepting the zero product $p=0$ would include an undefined angle at a zero radius vector.

Worked example

Eliminate parameters and prove the entire midpoint locus

Let $T$ be the intersection of the tangents at $A,B$, and $N$ the intersection of their normals. Since $p=-4$, the parameters are distinct. Subtract the two tangent equations: $(t_1-t_2)y=a(t_1^2-t_2^2)$. Dividing by the nonzero difference gives $y=as$, and substitution gives

$$T=(ap,as)=(-4a,as).$$

Subtracting the normal equations similarly gives $x=a(2+t_1^2+t_1t_2+t_2^2)=a(2+s^2-p)$. Substitution into either normal then gives $y=-aps$. Therefore

$$N=(a(s^2+6),4as),\qquad M=\frac{T+N}{2}=\left(a+\frac a2s^2,\frac{5a}{2}s\right).$$

Eliminating $s$ yields

$$y^2=\frac{25a}{2}(x-a).$$

This proves every constructed midpoint lies on the stated parabola. For the converse, take any point on it and set $s=2y/(5a)$. The quadratic $z^2-sz-4=0$ has discriminant $s^2+16\gt0$, so it gives distinct real, nonzero parameters with sum $s$ and product $-4$. Their radius vectors are nonzero and orthogonal, and the construction returns the given midpoint. Thus the locus is the whole parabola, not merely a subset of it.

Three normals and a fixed centroid

Theorem

The centroid when three distinct real normal contacts exist

Fix $a\gt0$ and a real $h$. For a point $H=(h,k)$, assume three distinct real contact parameters $t_1,t_2,t_3$ give normals through $H$. The centroid of the three contact points is

$$G=\left(\frac{2(h-2a)}3,0\right).$$

Thus it is fixed as $k$ varies among positions for which the stated three-contact assumption holds.

Substituting $H$ into the normal equation gives

$$at^3+(2a-h)t-k=0.$$

Its roots are contact parameters, not the coordinates of the given point. Under the hypothesis, the three real roots are distinct. Vieta's relations give

$$\sum_i t_i=0,\qquad \sum_{i\lt j}t_it_j=\frac{2a-h}{a},\qquad \sum_i t_i^2=\left(\sum_i t_i\right)^2-2\sum_{i\lt j}t_it_j =\frac{2(h-2a)}a.$$

Average the coordinates $(at_i^2,2at_i)$ to obtain the displayed centroid. These three distinct contacts are not collinear. If a genuine line $Ax+By+C=0$ contained them, substitution of $(at^2,2at)$ would make all three parameters roots of $Aa t^2+2aBt+C$. This polynomial has degree at most two and is not identically zero, since the line coefficients are not all zero. It cannot have three distinct roots. Thus the assumed contacts really form a triangle, not a degenerate three-point configuration.

The term $k$ affects the cubic's constant coefficient but not these two symmetric sums, which explains why the centroid depends on $h$ alone. The condition is essential: a real cubic need not have three distinct real roots. Algebraic identities involving complex or repeated roots cannot be interpreted as three distinct real contact points.

Worked example

A valid three-normal configuration and its boundary warning

Take $h=3a$, $k=0$. The normal equation becomes $at(t^2-1)=0$, giving $t=-1,0,1$. The contacts are $(a,-2a),(0,0),(a,2a)$, with centroid $(2a/3,0)$, exactly the theorem's value. Their normal lines have distinct directions and all pass through $(3a,0)$. The vertex contact is allowed here: unlike the preceding radius-angle problem, this construction does not require an angle from $O$ to that contact.

At $h=0,k=0$, the equation is $at(t^2+2)=0$, which has only one real root. There is then no triangle of three real contacts to which the centroid statement could apply. The result does not assert that every point on a vertical line supplies three normals.

Quick checks

Checkpoint

Q1. Why does squaring the focus-directrix equation give an equivalent locus here?

Check the converse and the signs of the quantities whose squares agree.

Solution · Quick-check Q1

If $y^2=4ax$ with $a\gt0$, then $x\ge0$ and $(x-a)^2+y^2=(x+a)^2$. Taking nonnegative square roots recovers the two distances, with $|x+a|=x+a\gt0$.

Checkpoint

Q2. Does a vertical tangent mean the normal formula fails at the vertex?

Use the equations rather than dividing by a slope.

Solution · Quick-check Q2

No. At $t=0$ the tangent is $x=0$ and the normal is $y=0$. The derivative $(0,2a)$ is nonzero, so the tangent direction is well defined.

Summary

The focus-directrix condition and $y^2=4ax$ are equivalent when the distance signs and $a\gt0$ are retained. A regular parametrization gives tangent and normal equations valid even at the vertex. Tangent counts depend on real quadratic roots; reflection follows by reversing a normal component. For locus and centroid arguments, real contacts, excluded degeneracies, and the reverse construction determine the scope of the conclusion.

Exercises

1. For $(x-1)^2=-12(y+2)$, identify the vertex, focus, directrix, and tangent and normal at the vertex.
2. Find both tangents from $H=(-a,0)$ to $y^2=4ax$, and their chord of contact. Keep $a\gt0$ and avoid dividing by the second coordinate of $H$.
3. In the perpendicular-radius construction, take $t_1=2$, $t_2=-2$. Calculate $A,B,T,N,M$ and verify the right-angle condition.
4. Suppose three distinct real normals through $(4a,k)$ exist. Find the contact centroid and explain precisely what the assumption contributes.

Guided solutions

Solution · Solution 1

Here $a=3$, the opening is downward, and the vertex is $(1,-2)$. The focus is $(1,-5)$, the directrix is $y=1$, the tangent at the vertex is $y=-2$, and the normal is $x=1$.

Solution · Solution 2

The parameter equation is $at^2-a=0$, so $t=1,-1$. The tangents are $x-y+a=0$ and $x+y+a=0$; the contacts are $(a,2a)$ and $(a,-2a)$. Their chord is $2a(x-a)=0$, or $x=a$.

Solution · Solution 3

$A=(4a,4a)$ and $B=(4a,-4a)$ are nonzero, with dot product $16a^2-16a^2=0$. Here $s=0,p=-4$, so $T=(-4a,0)$, $N=(6a,0)$, and $M=(a,0)$, the vertex of the midpoint locus.

Solution · Solution 4

The centroid is $(4a/3,0)$. The hypothesis supplies three distinct real contact parameters, so the Vieta sums represent a geometric centroid of three contacts. It does not hold merely because the point lies on $x=4a$; values of $k$ without three such contacts are outside the assertion.