Tangents, normals and the general equation of a conic

What you will be able to do

• write the tangent and the normal to an ellipse and a hyperbola, in point, parametric and slope forms;
• use the director circle and the properties of the foci and the tangent;
• say which conic a general second degree equation represents.

A mirror shaped like an ellipse

A room with a ceiling in the shape of half an ellipse has a curious property. If you whisper at one focus, a person standing at the other focus hears you clearly, even across the whole room. Every sound that leaves one focus and hits the ceiling bounces to the other focus. The reason is a property of the tangent, which we meet in this lesson. Then we turn to the question of recognising a conic from its equation.

The tangent at a point

For the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ and a point $(x_1, y_1)$ on it, the tangent is

For the hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ it is

This is the same rule as for the circle and the parabola: replace $x^2$ by $xx_1$ and $y^2$ by $yy_1$. In the parametric forms it is $\dfrac{x\cos\theta}{a} + \dfrac{y\sin\theta}{b} = 1$ for the ellipse, and $\dfrac{x\sec\theta}{a} - \dfrac{y\tan\theta}{b} = 1$ for the hyperbola.

Example. The point $\left(4, \dfrac{9}{5}\right)$ is on $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$, because $\dfrac{16}{25} + \dfrac{81}{25 \times 9} = \dfrac{16}{25} + \dfrac{9}{25} = 1$. The tangent is $\dfrac{4x}{25} + \dfrac{9y/5}{9} = 1$, which is $4x + 5y = 25$. Its slope is $-0.8$, and it meets the axes at $(6.25, 0)$ and $(0, 5)$.

Example. The point $\left(5, \dfrac{16}{3}\right)$ is on $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$. The tangent is $\dfrac{5x}{9} - \dfrac{y}{3} = 1$, which is $5x - 3y = 9$.

The normal

The normal at $(x_1, y_1)$ is perpendicular to the tangent. For the ellipse, the tangent has the normal direction $\left(\dfrac{x_1}{a^2}, \dfrac{y_1}{b^2}\right)$. So the normal has the direction of that vector, and its equation is

For the ellipse at $\left(4, \dfrac{9}{5}\right)$ the normal is $\dfrac{25x}{4} - 5y = 16$, with slope $\dfrac{5}{4} = 1.25$. The slopes of the tangent and normal, $-\dfrac{4}{5}$ and $\dfrac{5}{4}$, multiply to $-1$.

The tangent with a given slope

Let $y = mx + c$ be a tangent to the ellipse. Put $y = mx + c$ in the equation and demand a double root. The condition is

For the hyperbola it is

So the tangents with slope $m$ are $y = mx \pm\sqrt{a^2m^2 + b^2}$ for the ellipse and $y = mx \pm\sqrt{a^2m^2 - b^2}$ for the hyperbola. For the hyperbola a tangent with slope $m$ exists only if $a^2m^2 > b^2$, which means $|m| > \dfrac{b}{a}$, steeper than the asymptotes.

Example. The tangents of slope $1$ to $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$ are $y = x \pm\sqrt{16 + 9} = x \pm 5$. The tangents of slope $2$ to $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$ are $y = 2x \pm\sqrt{36 - 16} = 2x \pm 2\sqrt{5}$. A line of slope $1$ cannot touch this hyperbola, because $9 - 16 < 0$.

The director circle and the foci

Perpendicular tangents to a circle meet on a bigger circle. The same holds for these conics. The point of meeting lies on the director circle:

For $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$ the director circle is $x^2 + y^2 = 41$, with radius $\sqrt{41} \approx 6.40$. For $\dfrac{x^2}{25} - \dfrac{y^2}{16} = 1$ it is $x^2 + y^2 = 9$, with radius $3$. For a rectangular hyperbola $a = b$ the director circle shrinks to a point, the centre. For a parabola the director "circle" is the directrix.

Two more properties of the foci follow from the tangent equation.

• The product of the perpendicular distances from the two foci to any tangent is $b^2$, for both the ellipse and the hyperbola. The simulation shows $d_1 \times d_2 = b^2$.
• The tangent at $P$ makes equal angles with the focal lines $PF_1$ and $PF_2$ for the ellipse (it bisects the exterior angle at $P$), and the normal bisects the angle $F_1PF_2$. So a ray from one focus reflects through the other. For the hyperbola, the tangent bisects the angle $F_1PF_2$, so a ray aimed at one focus reflects towards the other.

Which conic is it?

The general second degree equation is

Two numbers decide the kind. The first is

If $\Delta = 0$, the equation is degenerate: it describes a pair of lines, a single point, or no real points. A pair of lines is a conic in the cone picture too, since a plane through the apex of the cone cuts it in lines. If $\Delta \ne 0$, the equation is a real conic, and the sign of $h^2 - ab$ names it:

• $h^2 - ab < 0$: an ellipse. It is a circle if $a = b$ and $h = 0$.
• $h^2 - ab = 0$: a parabola.
• $h^2 - ab > 0$: a hyperbola. It is rectangular if $a + b = 0$.

Why does $a + b = 0$ give a rectangular hyperbola? The sum $a + b$ does not change when the axes are turned. A rectangular hyperbola with its axes along the coordinate axes is $x^2 - y^2 = k$, and there $a + b = 1 - 1 = 0$.

Here are four examples.

1. $4x^2 + 9y^2 - 16x + 18y - 11 = 0$ has $h^2 - ab = -36 < 0$ and $\Delta = -1296 \ne 0$. It is an ellipse. Complete the squares: $4(x - 2)^2 + 9(y + 1)^2 = 36$, so $\dfrac{(x - 2)^2}{9} + \dfrac{(y + 1)^2}{4} = 1$. The centre is $(2, -1)$, $a = 3$, $b = 2$ and $c = \sqrt{5}$.
2. $x^2 + 2xy + y^2 - 8x + 8y = 0$ has $h^2 - ab = 1 - 1 = 0$ and $\Delta = -64 \ne 0$. It is a parabola, and the equation is $(x + y)^2 = 8(x - y)$.
3. $x^2 - 4y^2 - 2x + 16y - 19 = 0$ has $h^2 - ab = 4 > 0$ and $\Delta = 16 \ne 0$. It is a hyperbola: $\dfrac{(x - 1)^2}{4} - (y - 2)^2 = 1$ with centre $(1, 2)$.
4. $x^2 - y^2 - 2x + 2y = 0$ has $\Delta = 0$. It factors as $(x - y)(x + y - 2) = 0$, which is a pair of lines.

Practise with new numbers

Try it in Python

The program takes the six numbers of a general second degree equation. It works out $h^2 - ab$ and $\Delta$ and names the conic.

Common mistakes

• Putting $2h$ in the test as $h$. In $ax^2 + 2hxy + by^2$ the coefficient of $xy$ is $2h$. So $h$ is half of it.
• Forgetting $\Delta$. A pair of lines can have $h^2 - ab > 0$ and still not be a hyperbola.
• Using the ellipse rule for the hyperbola. The slope form has $a^2m^2 + b^2$ for an ellipse and $a^2m^2 - b^2$ for a hyperbola.
• Forgetting that some slopes give no tangent. A hyperbola has tangents only for slopes steeper than the asymptotes.

Remember these

• Tangent at $(x_1, y_1)$: $\dfrac{xx_1}{a^2} \pm \dfrac{yy_1}{b^2} = 1$. Slope form: $c^2 = a^2m^2 \pm b^2$.
• Director circle: $x^2 + y^2 = a^2 + b^2$ (ellipse) and $a^2 - b^2$ (hyperbola).
• The foci are at distances $d_1$ and $d_2$ from a tangent with $d_1d_2 = b^2$.
• $\Delta \ne 0$ and $h^2 - ab$ negative, zero or positive gives an ellipse, a parabola or a hyperbola.

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 $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$ at $\left(2, \dfrac{3\sqrt{3}}{2}\right)$.

2. For which values of $c$ is $y = x + c$ a tangent to $\dfrac{x^2}{4} + y^2 = 1$?

3. Find the director circle of $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$.

4. What conic is $x^2 + 4xy + 4y^2 + x = 0$?

5. The distances from the two foci of an ellipse to a tangent are $d_1$ and $d_2$. What is the value of $d_1d_2$?

References

• NCERT, Mathematics for Class XI, Chapter 10 (Conic Sections), the sections on the ellipse and the hyperbola. The general equation test and the director circle belong to the JEE Main syllabus, not the NCERT text.
• OpenStax, Precalculus 2e, Chapter 12 (Analytic Geometry), the sections on the ellipse, the hyperbola and rotation of axes.