Tangents and normals to a parabola

What you will be able to do

• write the tangent and the normal at a point of a parabola in the point, parametric and slope forms;
• find the tangents from an outside point and the chord of contact;
• use the fact that perpendicular tangents meet on the directrix;
• count the normals through a point.

A road touching a hill

A road that just touches the side of a parabolic hill is a tangent. A signpost standing straight out from the hill at that point lies along the normal. In this lesson we find both with algebra. The method is the same as for the circle: a tangent meets the curve once, so the algebra has a double root.

The tangent at a point

For $y^2 = 4ax$ the tangent at $(x_1, y_1)$ on the curve is

The rule is the same as for the circle: replace $y^2$ by $yy_1$, and $2x$ by $x + x_1$. Here $y^2 = 4ax$ becomes $yy_1 = 2a(x + x_1)$.

For the parabola $y^2 = 8x$, $a = 2$, and the point $(2, 4)$ is on it because $16 = 8 \times 2$. The tangent is $4y = 4(x + 2)$, which is $y = x + 2$.

In terms of the parameter, the point is $(at^2, 2at)$. Put $x_1 = at^2$ and $y_1 = 2at$ into the rule. You get $2aty = 2a(x + at^2)$, which is

The tangent with a given slope

Let the line $y = mx + c$ touch $y^2 = 4ax$. Put $y = mx + c$ in the equation: $(mx + c)^2 = 4ax$, which is $m^2x^2 + (2mc - 4a)x + c^2 = 0$. A tangent gives a double root, so the discriminant is zero: $(2mc - 4a)^2 - 4m^2c^2 = 0$. This simplifies to $16a^2 - 16amc = 0$, so

The tangent with slope $m$ is $y = mx + \dfrac{a}{m}$, and the line $y = mx + c$ is a tangent exactly when $c = \dfrac{a}{m}$. The point of contact is $\left(\dfrac{a}{m^2}, \dfrac{2a}{m}\right)$, which is the point $t = \dfrac{1}{m}$. This form does not give a vertical tangent, because $m$ cannot be infinite. The vertical tangent is $x = 0$ at the vertex.

Example. The tangent to $y^2 = 8x$ with slope $2$ is $y = 2x + 1$. It touches the parabola at $\left(\tfrac{1}{2}, 2\right)$, which is the point $t = \tfrac{1}{2}$. The slope of the tangent at $t$ is $\dfrac{1}{t} = 2$.

The normal

The normal at $P$ is the line through $P$ at right angles to the tangent. The tangent at $(x_1, y_1)$ has slope $\dfrac{2a}{y_1}$, so the normal has slope $-\dfrac{y_1}{2a}$ and

In the parametric form the tangent slope is $\dfrac{1}{t}$ and the normal slope is $-t$:

In the slope form, if the normal has slope $m$, then $t = -m$ and the normal is

and it meets the parabola at $(am^2, -2am)$.

For $y^2 = 8x$ at $(2, 4)$, the normal slope is $-\dfrac{4}{4} = -1$, so $y - 4 = -(x - 2)$, which is $x + y = 6$. The tangent $y = x + 2$ has slope $1$. The product of the slopes is $-1$, as it must be.

The subtangent and the subnormal

Let $P$ be a point on the parabola, $N$ the foot of its perpendicular on the axis, $T$ where the tangent meets the axis and $G$ where the normal meets the axis. The tangent $ty = x + at^2$ meets $y = 0$ at $x = -at^2$, so $T = (-at^2, 0)$. The normal meets the axis at $x = 2a + at^2$, so $G = (2a + at^2, 0)$. Then $N = (at^2, 0)$ and

The subnormal is the same for every point of a parabola. This is a special property of the parabola.

Tangents from an outside point

From a point $(x_1, y_1)$ outside the parabola, two tangents can be drawn. Write the tangent $y = mx + \dfrac{a}{m}$ and put the point in it. This gives a quadratic in $m$. Its two roots are the slopes of the two tangents. The line through the two points of contact is the chord of contact, $yy_1 = 2a(x + x_1)$, the same rule as the tangent. The pair of tangents is $SS_1 = T^2$.

Example. Take the parabola $y^2 = 4x$ and the point $(-1, 3)$. The tangent $y = mx + \dfrac{1}{m}$ passes through it when $3 = -m + \dfrac{1}{m}$, which is $m^2 + 3m - 1 = 0$. The roots are $m = \dfrac{-3 \pm \sqrt{13}}{2}$. They multiply to $-1$, so the tangents are perpendicular.

That was no accident. The point $(-1, 3)$ is on the directrix $x = -1$. For a point $(-a, k)$ on the directrix of $y^2 = 4ax$ the quadratic is $m^2 + \dfrac{k}{a}m - 1 = 0$. The product of the roots is always $-1$. So perpendicular tangents meet on the directrix, and the directrix is the locus of the point from which the tangents are at right angles. It is the parabola's counterpart of the director circle.

How many normals?

Suppose a normal $y = mx - 2am - am^3$ passes through $(h, k)$. Then $k = mh - 2am - am^3$, which is

This is a cubic in $m$, so there are up to three normals through any point. For example, take $y^2 = 4x$ and $(5, 0)$. Here $a = 1$, $h = 5$ and $k = 0$, so $m^3 - 3m = 0$ and $m = 0$ or $m = \pm\sqrt{3}$. The normal with $m = 0$ is the axis. The other two touch the parabola at $(am^2, -2am) = (3, \mp 2\sqrt{3})$. Through a point on the axis with $h > 2a$ there are three normals, and for $h \le 2a$ only the axis.

Practise with new numbers

Try it in Python

The program takes a point of the parabola by its parameter $t$. It finds the tangent and the normal and checks that the tangent has only one point in common with the parabola.

Common mistakes

• Writing the tangent slope as $m = \dfrac{y_1}{2a}$. The slope of the tangent is $\dfrac{2a}{y_1}$.
• Mixing the tangent with the normal. The normal has slope $-t$, the tangent has slope $\dfrac{1}{t}$.
• Forgetting the case $m = 0$ for normals. The axis is itself a normal at the vertex.
• Dropping the vertical tangent. The slope form cannot give $x = 0$.

Remember these

• Tangent at $(x_1, y_1)$: $yy_1 = 2a(x + x_1)$. At $(at^2, 2at)$: $ty = x + at^2$.
• Tangent with slope $m$: $y = mx + \dfrac{a}{m}$, touching at $\left(\dfrac{a}{m^2}, \dfrac{2a}{m}\right)$.
• Normal at $(at^2, 2at)$: $y + tx = 2at + at^3$. With slope $m$: $y = mx - 2am - am^3$.
• The subtangent is $2at^2$ and the subnormal is $2a$.
• Perpendicular tangents meet on the directrix. Up to three normals pass through a point.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the tangent to $y^2 = 4x$ at $(1, 2)$.

2. Find the tangent with slope $1$ to $y^2 = 12x$.

3. Find the normal to $y^2 = 4x$ at the point with $t = 2$, and where it meets the axis.

4. How many normals pass through $(3, 0)$ for $y^2 = 4x$?

5. Two perpendicular tangents to $y^2 = 12x$ meet at a point whose $x$-coordinate is not given. What is it?

References

• NCERT, Mathematics for Class XI, Chapter 10 (Conic Sections), the section on the parabola.
• OpenStax, Calculus Volume 3, Chapter 7 (Parametric Equations and Polar Coordinates), the section on conic sections.