Hyperbola — Parametric and Auxiliary Circle

If you have read the article on the ellipse's auxiliary circle, you know the trick: take a circle with the same centre and radius a, pick a point on it at angle \theta, and project down to the ellipse. The result is the clean parametric form (a\cos\theta, b\sin\theta).

The hyperbola needs the same kind of tool. The equation \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 is implicit — you cannot easily write y as a single function of x (there are two branches, each with a square root). A parametric form gives you a single parameter \theta that traces the entire curve, with both coordinates expressed as clean trigonometric functions.

But there is a twist. The ellipse parametrisation uses \cos\theta and \sin\theta, which satisfy \cos^2\theta + \sin^2\theta = 1 — matching the plus sign in the ellipse equation. The hyperbola has a minus sign: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1. The identity that has a minus sign is \sec^2\theta - \tan^2\theta = 1. So the natural parametrisation of a hyperbola uses \sec\theta and \tan\theta, not \cos\theta and \sin\theta.

The auxiliary circle

The auxiliary circle passes through the vertices (\pm a, 0) of the hyperbola. Unlike the ellipse case (where the circle entirely encloses the ellipse), the auxiliary circle is smaller than the hyperbola in the y-direction — the hyperbola's branches extend far beyond the circle.

The purpose of the auxiliary circle is the same as for the ellipse: to provide an angular coordinate system for points on the hyperbola.

The eccentric angle and parametric form

Here is how the projection works. Take a point Q on the auxiliary circle at angle \theta from the positive x-axis: Q = (a\cos\theta, a\sin\theta). Drop a vertical line through Q — that is, look at the vertical line x = a\cos\theta.

For this vertical line to intersect the hyperbola, you need |x| \geq a, which means |\cos\theta| \geq 1. But |\cos\theta| \leq 1 always, with equality only at \theta = 0 and \theta = \pi. So the direct vertical projection that works for the ellipse does not work for the hyperbola (except at the vertices).

Instead, the connection goes through the tangent line. Draw the tangent to the auxiliary circle at Q. This tangent is perpendicular to the radius CQ and meets the x-axis at some point N. Then draw a vertical line through N. This vertical line intersects the hyperbola at the corresponding point P.

Let's compute. The tangent to x^2 + y^2 = a^2 at Q = (a\cos\theta, a\sin\theta) has equation

This tangent meets the x-axis (y = 0) at x\cos\theta = a, so x = \frac{a}{\cos\theta} = a\sec\theta. The point N = (a\sec\theta, 0).

Now, the point P on the hyperbola has x-coordinate a\sec\theta. Substituting into \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1:

Taking the sign consistent with the quadrant of \theta: y = b\tan\theta.

Verification. Substitute x = a\sec\theta and y = b\tan\theta into the hyperbola equation:

The identity \sec^2\theta - \tan^2\theta = 1 does the work, just as \cos^2\theta + \sin^2\theta = 1 does for the ellipse.

Excluded values. At \theta = \pi/2 and \theta = 3\pi/2, both \sec\theta and \tan\theta are undefined. These angles correspond to the "top" and "bottom" of the auxiliary circle — points where the tangent to the circle is vertical and does not meet the x-axis.

Which branch? When \theta \in (-\pi/2, \pi/2), \sec\theta > 0, so x > 0: the point is on the right branch. When \theta \in (\pi/2, 3\pi/2), \sec\theta < 0, so x < 0: the point is on the left branch.

Key angles and the points they give

Notice that as \theta \to \pi/2, both \sec\theta and \tan\theta tend to +\infty, so the point moves out along the right branch toward infinity. The asymptote y = \frac{b}{a}x corresponds to the direction as \theta \to \pi/2, since \frac{b\tan\theta}{a\sec\theta} = \frac{b\sin\theta}{a} \to \frac{b}{a}.

Comparison with the ellipse parametrisation

The parallel structure is not a coincidence. The hyperbolic functions (\cosh and \sinh) provide yet another parametrisation — (\alpha\cosh t, b\sinh t) — using \cosh^2 t - \sinh^2 t = 1. That parametrisation is smooth and has no excluded values, but it is less commonly used in Indian textbooks.

The director circle

When you draw two tangent lines from an external point to the hyperbola, those tangents generally meet at some angle. The special points where the two tangents are perpendicular — meeting at exactly 90° — form a circle called the director circle.

Derivation

The tangent to the hyperbola in slope form is y = mx \pm \sqrt{a^2 m^2 - b^2} (this comes from the condition for a line y = mx + c to touch the hyperbola, which gives c^2 = a^2 m^2 - b^2).

If a point (h, k) lies on both tangents, then k - mh = \pm\sqrt{a^2 m^2 - b^2}. Squaring:

This is a quadratic in m. Let m_1 and m_2 be its roots — the slopes of the two tangents. By Vieta's formulas:

For perpendicular tangents, m_1 m_2 = -1:

Replacing (h, k) with (x, y): x^2 + y^2 = a^2 - b^2.

Three cases

The radius squared of the director circle is a^2 - b^2. This leads to three cases:

• a > b: The director circle exists with radius \sqrt{a^2 - b^2}. Points inside this circle can send two perpendicular tangents to the hyperbola.

• a = b (rectangular hyperbola): a^2 - b^2 = 0. The "circle" degenerates to a single point — the centre. The only point from which perpendicular tangents can be drawn to a rectangular hyperbola is its centre.

• a < b: a^2 - b^2 < 0. No real circle. There is no point from which two perpendicular tangents to the hyperbola exist.

Compare this with the ellipse, where the director circle is x^2 + y^2 = a^2 + b^2 — always positive, always exists. The minus sign in the hyperbola equation turns the plus in a^2 + b^2 into a minus in a^2 - b^2, and that minus sign is enough to make the director circle fail for some hyperbolas.

The latus rectum

The latus rectum of a conic is the chord through a focus perpendicular to the major (or transverse) axis. For the hyperbola, it gives you a measure of how "wide" the curve is at the level of the foci.

Derivation

The foci are at (\pm c, 0) where c^2 = a^2 + b^2. The latus rectum through the focus (c, 0) is the vertical line x = c. Substitute x = c into the hyperbola equation:

(using c^2 - a^2 = b^2)

The two endpoints of the latus rectum are \left(c, \frac{b^2}{a}\right) and \left(c, -\frac{b^2}{a}\right). The full length is

This formula is identical to the latus rectum of an ellipse — the same expression 2b^2/a appears for both curves. The difference is that for an ellipse b < a (so the latus rectum is shorter than 2a), while for a hyperbola there is no such constraint — b can be larger than a.

The latus rectum in terms of eccentricity

Since b^2 = a^2(e^2 - 1) for a hyperbola (from c^2 = a^2 + b^2 and e = c/a):

For the rectangular hyperbola (e = \sqrt{2}), this gives \ell = 2a(2 - 1) = 2a, which equals the transverse axis. So for a rectangular hyperbola, the latus rectum is exactly as long as the transverse axis — an elegant coincidence.

Worked examples

Common confusions

• "The parametric form is (a\cos\theta, b\sin\theta), same as the ellipse." That is the ellipse parametrisation. The hyperbola uses (a\sec\theta, b\tan\theta). The identity driving the hyperbola is \sec^2\theta - \tan^2\theta = 1, not \cos^2\theta + \sin^2\theta = 1. Confusing them will produce points that do not lie on the hyperbola.

• "\theta is the angle that CP makes with the x-axis." No. \theta is the eccentric angle — the angle on the auxiliary circle, not the angle to the point on the hyperbola. The actual angle from C to P is \arctan\left(\frac{b\tan\theta}{a\sec\theta}\right) = \arctan\left(\frac{b\sin\theta}{a}\right), which is different from \theta (unless a = b and the hyperbola is rectangular).

• "The director circle always exists." It exists only when a^2 > b^2, i.e., a > b. When a = b, it degenerates to a point. When a < b, no point in the plane sends perpendicular tangents to the hyperbola. For the ellipse, the director circle always exists because the formula has a^2 + b^2 (always positive), not a^2 - b^2.

• "The latus rectum formula is 2a^2/b." The correct formula is 2b^2/a, where a is the semi-transverse axis and b is the semi-conjugate axis. Swapping a and b gives the wrong answer. The same formula applies to ellipses and parabolas — it is universal across conics.

• "The auxiliary circle is x^2 + y^2 = b^2." No. The auxiliary circle uses a, the semi-transverse axis (the larger denominator in the hyperbola equation), not b. Using b would give a circle smaller than the distance between the vertices, which would not pass through them.

Going deeper

Parametric form of xy = c^2

The rectangular hyperbola xy = c^2 has its own parametric form, even simpler than the general case:

Verification: xy = ct \cdot \frac{c}{t} = c^2. The parameter t is not an angle — it is a real-valued scaling parameter. As t \to 0^+, the point moves up the y-axis; as t \to \infty, it moves along the x-axis toward the right.

The slope of the tangent at the point (ct, c/t) is dy/dx = -1/t^2 (derived in the conjugate and rectangular article). This means the tangent becomes flatter as t grows and steeper as t shrinks — consistent with the shape of the curve.

The equation of the chord joining two points (ct_1, c/t_1) and (ct_2, c/t_2) on xy = c^2 is

This is a clean formula that avoids fractions, making it a favourite tool in competitive mathematics.

Alternate derivation of the director circle using the discriminant

The tangent to \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 with slope m is y = mx + \sqrt{a^2 m^2 - b^2} (or minus the root). For this tangent to be real, you need a^2 m^2 - b^2 \geq 0, i.e., |m| \geq b/a. This already tells you that tangent lines to a hyperbola cannot have arbitrary slopes — slopes between -b/a and b/a are forbidden. (Those are the slopes of lines that would pass between the asymptotes without touching the curve.)

Now, if this tangent passes through (h, k): (k - mh)^2 = a^2 m^2 - b^2, giving

For two tangent lines that are perpendicular, m_1 m_2 = -1. By Vieta's formulas, m_1 m_2 = \frac{k^2 + b^2}{h^2 - a^2} = -1. This gives h^2 + k^2 = a^2 - b^2.

But there is an additional constraint: both roots must be real and satisfy |m| \geq b/a. For the discriminant of the quadratic to be non-negative:

Expanding: 4k^2 h^2 - 4(h^2 k^2 + h^2 b^2 - a^2 k^2 - a^2 b^2) \geq 0

Using h^2 + k^2 = a^2 - b^2 (the director circle condition): h^2 = a^2 - b^2 - k^2. Substituting:

This is always true (sum of non-negative terms). So the discriminant condition is automatically satisfied whenever the director circle condition holds — confirming that the director circle equation is both necessary and sufficient.

The relation \ell = a(e^2 - 1) \cdot 2/e \cdot e

Bhaskara II, in his Siddhanta Shiromani, systematically studied the properties of conic-like curves through their focal chords. The latus rectum can be expressed as \ell = 2a(e^2 - 1), which for a rectangular hyperbola (e = \sqrt{2}) gives \ell = 2a — the latus rectum equals the transverse axis. This special property makes the rectangular hyperbola's geometry particularly clean: the semi-latus rectum b^2/a = a is the same as the semi-transverse axis.

Where this leads next

• Hyperbola -- Tangent and Normal — equations of tangent and normal at a point given in parametric form, including the tangent at (a\sec\theta, b\tan\theta).
• Hyperbola -- Introduction — the standard equation, eccentricity, and all the basic terms, if you need a refresher.
• Ellipse -- Auxiliary Circle and Eccentric Angle — the ellipse version of the same construction, for comparison.
• Trigonometric Ratios — the \sec\theta and \tan\theta functions used in the parametric form, if you need to brush up.
• Circle -- Standard Forms — the auxiliary and director circles are plain circles. This article covers their equations and properties.