Hyperbola: Chord, Tangent And Normal

Q: What is the equation of a tangent to a hyperbola in slope form?

For x^2/a^2 - y^2/b^2 = 1, the tangents of slope m are y = mx plus or minus sqrt(a^2 m^2 - b^2). They are real only when |m| > b/a, and the point of contact is (-a^2 m/c, -b^2/c), where c is the chosen intercept.

Q: What is the condition for a line to touch a hyperbola?

The line y = mx + c touches x^2/a^2 - y^2/b^2 = 1 if and only if c^2 = a^2 m^2 - b^2. For a line written as lx + my + n = 0, the equivalent condition is a^2 l^2 - b^2 m^2 = n^2.

Q: What is the equation of the normal to a hyperbola?

At (x1, y1) the normal is a^2 x/x1 + b^2 y/y1 = a^2 + b^2. At the parametric point (a sec(theta), b tan(theta)) it is a x cos(theta) + b y cot(theta) = a^2 + b^2. In slope form it is y = mx plus or minus (a^2 + b^2)m / sqrt(a^2 - b^2 m^2).

Q: What is the reflection property of a hyperbola?

The tangent at any point P bisects the angle between the focal radii SP and S'P. So a ray from one focus is reflected as if it came from the other focus, and a ray aimed at one focus from outside is reflected towards the other focus. This is why confocal ellipses and hyperbolas cut at right angles.

MathsHyperbolaFor JEE aspirants

The tangent to the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ at $(x_{1}, y_{1})$ is $\frac{x x _{1}}{a ^{2}} - \frac{y y _{1}}{b ^{2}} = 1$ , and a line $y = m x + c$ touches the hyperbola when $c^{2} = a^{2} m^{2} - b^{2}$ . The normal at $(x_{1}, y_{1})$ is $\frac{a ^{2} x}{x _{1}} + \frac{b ^{2} y}{y _{1}} = a^{2} + b^{2}$ . This page on hyperbola chord, tangent and normal covers all three tangent forms, normals, pair of tangents, director circle, chord of contact, chord with a given midpoint, diameters and the reflection property, with solved examples for JEE Main and JEE Advanced.

Key Formulas - Quick Reference

Chord joining $P (α)$ and $Q (β)$ : $\frac{x}{a} cos \frac{α - β}{2} - \frac{y}{b} sin \frac{α + β}{2} = cos \frac{α + β}{2}$
$y = m x + c$ is a secant, tangent or misses the curve according as $c^{2} >, =, < a^{2} m^{2} - b^{2}$
Tangent, slope form: $y = m x \pm a^{2} m^{2} - b^{2}$ , touching at $(- \frac{a ^{2} m}{c}, - \frac{b ^{2}}{c})$
Tangent, point form: $\frac{x x _{1}}{a ^{2}} - \frac{y y _{1}}{b ^{2}} = 1$ ; parametric form: $\frac{x sec θ}{a} - \frac{y tan θ}{b} = 1$
Normal: $\frac{a ^{2} x}{x _{1}} + \frac{b ^{2} y}{y _{1}} = a^{2} + b^{2}$ ; at $θ$ : $\frac{a x}{sec θ} + \frac{b y}{tan θ} = a^{2} + b^{2}$
Normal, slope form: $y = m x \pm \frac{( a ^{2} + b ^{2} ) m}{a ^{2} - b ^{2} m ^{2}}$
Pair of tangents $S S_{1} = T^{2}$ ; chord of contact $T = 0$ ; chord with midpoint $(x_{1}, y_{1})$ : $T = S_{1}$
Director circle: $x^{2} + y^{2} = a^{2} - b^{2}$
Diameter bisecting chords of slope $m$ : $y = \frac{b ^{2}}{a ^{2} m} x$

1. Chord Joining Two Points

The equation of the chord of the hyperbola joining the two points $P (α) \equiv (a sec α, b tan α)$ and $Q (β) \equiv (a sec β, b tan β)$ is

\frac{x}{a} cos \frac{α - β}{2} - \frac{y}{b} sin \frac{α + β}{2} = cos \frac{α + β}{2}

Notation used in this chapter

For the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ and a point $(x_{1}, y_{1})$ , we write

S \equiv \frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} - 1, S_{1} \equiv \frac{x _{1}^{2}}{a ^{2}} - \frac{y _{1}^{2}}{b ^{2}} - 1, T \equiv \frac{x x _{1}}{a ^{2}} - \frac{y y _{1}}{b ^{2}} - 1

2. Line and a Hyperbola

Substituting $y = m x + c$ into $b^{2} x^{2} - a^{2} y^{2} = a^{2} b^{2}$ gives the quadratic

(b^{2} - a^{2} m^{2}) x^{2} - 2 a^{2} m c x - a^{2} (c^{2} + b^{2}) = 0

Its discriminant simplifies to $4 a^{2} b^{2} (c^{2} - a^{2} m^{2} + b^{2})$ . So the straight line $y = m x + c$

is a secant (cuts in two points) if $c^{2} > a^{2} m^{2} - b^{2}$ ,
is a tangent if $c^{2} = a^{2} m^{2} - b^{2}$ ,
passes outside the hyperbola if $c^{2} < a^{2} m^{2} - b^{2}$ .

Remark: If $m = \pm \frac{b}{a}$ , the coefficient of $x^{2}$ vanishes and the equation becomes linear. A line parallel to an asymptote meets the hyperbola in at most one point, but it is not a tangent. For $c = 0$ it is the asymptote itself and never meets the curve.

3. Equation of Tangent to a Hyperbola

(i) Slope form

y = m x \pm a^{2} m^{2} - b^{2}

is a tangent to $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ having slope $m$ . Real tangents of slope $m$ exist only when $∣ m ∣ > \frac{b}{a}$ .

Point of contact: comparing $y = m x + c$ with the point form below gives the point where it touches:

(- \frac{a ^{2} m}{c}, - \frac{b ^{2}}{c}), where c = \pm a^{2} m^{2} - b^{2}

(ii) Point form

The equation of the tangent to $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ at the point $(x_{1}, y_{1})$ is

\frac{x x _{1}}{a ^{2}} - \frac{y y _{1}}{b ^{2}} = 1

(iii) Parametric form

The equation of the tangent to $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ at the point $(a sec θ, b tan θ)$ is

\frac{x sec θ}{a} - \frac{y tan θ}{b} = 1

Figure 1: Tangent

\frac{x sec θ}{a} - \frac{y tan θ}{b} = 1

and normal

\frac{a x}{sec θ} + \frac{b y}{tan θ} = a^{2} + b^{2}

P (a sec θ, b tan θ)

Note:

(i) The point of intersection of the tangents at $P (θ_{1})$ and $Q (θ_{2})$ is $(\frac{a cos \frac{θ _{1} - θ _{2}}{2}}{cos \frac{θ _{1} + θ _{2}}{2}}, b tan \frac{θ _{1} + θ _{2}}{2})$
(ii) If $∣ θ_{1} + θ_{2} ∣ = π$ , the tangents at $θ_{1}$ and $θ_{2}$ are parallel, since the slope of the tangent at $θ$ is $\frac{b}{a sin θ}$ .
(iii) There are two parallel tangents with the same slope $m$ . They touch the hyperbola at the extremities of a diameter.

Solved Example 1

Prove that the straight line

ℓ x + m y + n = 0

touches the hyperbola

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1

a^{2} ℓ^{2} - b^{2} m^{2} = n^{2}

Solution:

The given line is $y = - \frac{ℓ}{m} x - \frac{n}{m}$ . Comparing with $y = M x + c$ gives $M = - \frac{ℓ}{m}$ and $c = - \frac{n}{m}$ .

This line touches the hyperbola if $c^{2} = a^{2} M^{2} - b^{2}$ :

\frac{n ^{2}}{m ^{2}} = \frac{a ^{2} ℓ ^{2}}{m ^{2}} - b^{2} \Rightarrow a^{2} ℓ^{2} - b^{2} m^{2} = n^{2}

Hence proved. This is the condition of tangency $a^{2} ℓ^{2} - b^{2} m^{2} = n^{2}$ .

Solved Example 2

Find the equation of the tangent to the hyperbola

x^{2} - 4 y^{2} = 36

which is perpendicular to the line

x - y + 4 = 0

Solution:

The given line has slope 1. If $m$ is the slope of the tangent, then $m \times 1 = - 1$ , so $m = - 1$ .

Writing $x^{2} - 4 y^{2} = 36$ as $\frac{x ^{2}}{36} - \frac{y ^{2}}{9} = 1$ gives $a^{2} = 36$ and $b^{2} = 9$ . The tangents are

y = (- 1) x \pm 36 \times (- 1)^{2} - 9 = - x \pm 27

So the required tangents are $x + y \pm 33 = 0$ .

Solved Example 3

Find the equations and the length of the common tangents to the hyperbolas

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1

and

\frac{y ^{2}}{a ^{2}} - \frac{x ^{2}}{b ^{2}} = 1

Solution:

The tangent at $(a sec ϕ, b tan ϕ)$ on the first hyperbola is

\frac{x}{a} sec ϕ - \frac{y}{b} tan ϕ = 1 \dots (1)

A general point on the second hyperbola is $(b tan θ, a sec θ)$ , and the tangent there is

\frac{y}{a} sec θ - \frac{x}{b} tan θ = 1 \dots (2)

For a common tangent, (1) and (2) must be identical. Comparing the coefficients of $x$ and $y$ :

\frac{sec ϕ}{a} = - \frac{tan θ}{b} \dots (3) and - \frac{tan ϕ}{b} = \frac{sec θ}{a} \dots (4)

So $sec θ = - \frac{a}{b} tan ϕ$ and $tan θ = - \frac{b}{a} sec ϕ$ . Using $sec^{2} θ - tan^{2} θ = 1$ :

\frac{a ^{2}}{b ^{2}} tan^{2} ϕ - \frac{b ^{2}}{a ^{2}} (1 + tan^{2} ϕ) = 1 (\frac{a ^{2}}{b ^{2}} - \frac{b ^{2}}{a ^{2}}) tan^{2} ϕ = 1 + \frac{b ^{2}}{a ^{2}} tan^{2} ϕ = \frac{b ^{2}}{a ^{2} - b ^{2}}, sec^{2} ϕ = \frac{a ^{2}}{a ^{2} - b ^{2}}

These are real only when $a > b$ , so common tangents exist only if $a > b$ . Let $k = a^{2} - b^{2}$ and take $sec ϕ = \frac{a}{k}, tan ϕ = \frac{b}{k}$ . Then the points of contact are $(\frac{a ^{2}}{k}, \frac{b ^{2}}{k})$ on the first hyperbola and, from (3) and (4), $(- \frac{b ^{2}}{k}, - \frac{a ^{2}}{k})$ on the second. The other sign choices give the remaining tangents in the same way.

The length of the common tangent is the distance between the points of contact:

(\frac{a ^{2} + b ^{2}}{k})^{2} + (\frac{a ^{2} + b ^{2}}{k})^{2} = 2 \frac{a ^{2} + b ^{2}}{a ^{2} - b ^{2}}

Putting the values of $sec ϕ$ and $tan ϕ$ in (1) gives $\frac{x}{k} - \frac{y}{k} = 1$ . Taking all sign choices, the common tangents are $x \pm y = \pm a^{2} - b^{2}$ , each of length $2 \frac{a ^{2} + b ^{2}}{a ^{2} - b ^{2}}$ .

4. Equation of Normal to a Hyperbola

(a) Point form

The slope of the tangent at $(x_{1}, y_{1})$ is $\frac{b ^{2} x _{1}}{a ^{2} y _{1}}$ , so the normal has slope $- \frac{a ^{2} y _{1}}{b ^{2} x _{1}}$ . Simplifying $y - y_{1} = - \frac{a ^{2} y _{1}}{b ^{2} x _{1}} (x - x_{1})$ gives

\frac{a ^{2} x}{x _{1}} + \frac{b ^{2} y}{y _{1}} = a^{2} + b^{2} = a^{2} e^{2}

(b) Parametric form

The normal at $P (a sec θ, b tan θ)$ on the hyperbola is

\frac{a x}{sec θ} + \frac{b y}{tan θ} = a^{2} + b^{2} = a^{2} e^{2}

(c) Slope form

The normals with slope $m$ are

y = m x \pm \frac{( a ^{2} + b ^{2} ) m}{a ^{2} - b ^{2} m ^{2}}

Solved Example 4

A normal to the hyperbola

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1

meets the axes in

M

and

N

, and lines

M P

and

N P

are drawn perpendicular to the axes meeting at

P

. Prove that the locus of

P

is the hyperbola

a^{2} x^{2} - b^{2} y^{2} = (a^{2} + b^{2})^{2}

Solution:

The normal at $Q (a sec ϕ, b tan ϕ)$ is

a x cos ϕ + b y cot ϕ = a^{2} + b^{2} \dots (1)

Figure 2: The normal at

Q (a sec ϕ, b tan ϕ)

meets the axes at

M

and

N

. The perpendiculars

M P

and

N P

meet at

P

, whose locus is

a^{2} x^{2} - b^{2} y^{2} = (a^{2} + b^{2})^{2}

It meets the $x$ -axis at $M (\frac{a ^{2} + b ^{2}}{a} sec ϕ, 0)$ and the $y$ -axis at $N (0, \frac{a ^{2} + b ^{2}}{b} tan ϕ)$ .

The line $M P$ through $M$ perpendicular to the $x$ -axis is

x = \frac{a ^{2} + b ^{2}}{a} sec ϕ or sec ϕ = \frac{a x}{a ^{2} + b ^{2}} \dots (2)

The line $N P$ through $N$ perpendicular to the $y$ -axis is

y = \frac{a ^{2} + b ^{2}}{b} tan ϕ or tan ϕ = \frac{b y}{a ^{2} + b ^{2}} \dots (3)

Eliminating $ϕ$ from (2) and (3) using $sec^{2} ϕ - tan^{2} ϕ = 1$ :

\frac{a ^{2} x ^{2}}{( a ^{2} + b ^{2} ) ^{2}} - \frac{b ^{2} y ^{2}}{( a ^{2} + b ^{2} ) ^{2}} = 1

So the required locus of $P$ is $a^{2} x^{2} - b^{2} y^{2} = (a^{2} + b^{2})^{2}$ .

5. Pair of Tangents JEE Advanced level

The equation of the pair of tangents drawn from a point $(x_{1}, y_{1})$ to the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ is

S S_{1} = T^{2}

where $S$ , $S_{1}$ and $T$ are as defined in Section 1.

How many tangents? From a point outside the hyperbola that is not on an asymptote, two tangents can be drawn. From a point on an asymptote (other than the centre), only one tangent exists, because one of the lines given by $S S_{1} = T^{2}$ is the asymptote itself. No tangent can be drawn from the centre or from a point inside the curve.

Solved Example 5

How many real tangents can be drawn from the point

(4, 3)

to the hyperbola

\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1

? Find the equations of these tangents and the angle between them.

Solution:

Here $P \equiv (4, 3)$ and $S \equiv \frac{x ^{2}}{16} - \frac{y ^{2}}{9} - 1 = 0$ . Then $S_{1} = \frac{16}{16} - \frac{9}{9} - 1 = - 1 < 0$ , so $P$ lies outside the hyperbola.

The equation of the pair of lines $S S_{1} = T^{2}$ is

(\frac{x ^{2}}{16} - \frac{y ^{2}}{9} - 1) (- 1) = (\frac{4 x}{16} - \frac{3 y}{9} - 1)^{2} - \frac{x ^{2}}{16} + \frac{y ^{2}}{9} + 1 = \frac{x ^{2}}{16} + \frac{y ^{2}}{9} + 1 - \frac{x y}{6} - \frac{x}{2} + \frac{2 y}{3} 3 x^{2} - 4 x y - 12 x + 16 y = 0

This factorises as $(x - 4) (3 x - 4 y) = 0$ . The line $3 x - 4 y = 0$ , that is $y = \frac{3}{4} x$ , is an asymptote of the hyperbola and never touches it at a finite point. The point $(4, 3)$ lies on this asymptote, so only one genuine tangent exists.

Figure 3:

P (4, 3)

lies on the asymptote

3 x - 4 y = 0

\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1

. The only real tangent from

P

x = 4

The angle between the two lines of $3 x^{2} - 4 x y - 12 x + 16 y = 0$ is $θ = tan^{- 1} \frac{4}{3}$ , which is the angle between the tangent $x = 4$ and the asymptote.

Hence only one real tangent, $x = 4$ , can be drawn from $(4, 3)$ . It touches the hyperbola at the vertex $(4, 0)$ .

Solved Example 6

Find the locus of the point of intersection of perpendicular tangents to the hyperbola

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1

Solution:

Let $P (h, k)$ be the point of intersection of two perpendicular tangents. The pair of tangents from $P$ is $S S_{1} = T^{2}$ :

(\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} - 1) (\frac{h ^{2}}{a ^{2}} - \frac{k ^{2}}{b ^{2}} - 1) = (\frac{h x}{a ^{2}} - \frac{k y}{b ^{2}} - 1)^{2}

Collecting the $x^{2}$ and $y^{2}$ terms:

\frac{x ^{2}}{a ^{2}} (- \frac{k ^{2}}{b ^{2}} - 1) - \frac{y ^{2}}{b ^{2}} (\frac{h ^{2}}{a ^{2}} - 1) + \dots = 0 \dots (i)

Since (i) represents two perpendicular lines, the sum of the coefficients of $x^{2}$ and $y^{2}$ is zero:

\frac{1}{a ^{2}} (- \frac{k ^{2}}{b ^{2}} - 1) - \frac{1}{b ^{2}} (\frac{h ^{2}}{a ^{2}} - 1) = 0 \Rightarrow - k^{2} - b^{2} - h^{2} + a^{2} = 0

So the locus is $x^{2} + y^{2} = a^{2} - b^{2}$ .

6. Director Circle JEE Advanced level

The locus of the point of intersection of tangents that are at right angles is called the director circle of the hyperbola. Its equation is

x^{2} + y^{2} = a^{2} - b^{2}

Quick derivation: if the tangent $y = m x \pm a^{2} m^{2} - b^{2}$ passes through $(h, k)$ , then $m^{2} (h^{2} - a^{2}) - 2 hk m + (k^{2} + b^{2}) = 0$ . For perpendicular tangents the product of the roots is $- 1$ , so $\frac{k ^{2} + b ^{2}}{h ^{2} - a ^{2}} = - 1$ , giving $h^{2} + k^{2} = a^{2} - b^{2}$ .

Figure 4: Director circle

x^{2} + y^{2} = a^{2} - b^{2}

, drawn for

a = 2, b = 1

. The two tangents from any point

T

on it are perpendicular.

If $b^{2} < a^{2}$ , the director circle is real.
If $b^{2} = a^{2}$ (rectangular hyperbola), the radius is zero and the director circle reduces to the point circle at the centre. The only perpendicular pair of lines through the centre is the pair of asymptotes, so no two real tangents of this hyperbola are perpendicular.
If $b^{2} > a^{2}$ , the radius is imaginary, so there is no such circle and no pair of perpendicular tangents can be drawn to the curve.

7. Chord of Contact JEE Advanced level

The equation of the chord of contact of tangents drawn from a point $P (x_{1}, y_{1})$ to the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ is

T = 0, that is, \frac{x x _{1}}{a ^{2}} - \frac{y y _{1}}{b ^{2}} = 1

Solved Example 7

If tangents to the parabola

y^{2} = 4 a x

intersect the hyperbola

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1

A

and

B

, find the locus of the point of intersection of the tangents to the hyperbola at

A

and

B

Solution:

Let $P \equiv (h, k)$ be the point of intersection of the tangents at $A$ and $B$ . Then $A B$ is the chord of contact of $P$ :

\frac{x h}{a ^{2}} - \frac{y k}{b ^{2}} = 1 \dots (i)

This line touches the parabola $y^{2} = 4 a x$ . A tangent to the parabola is $y = m x + \frac{a}{m}$ , that is,

m x - y + \frac{a}{m} = 0 \dots (ii)

Lines (i) and (ii) must be the same, so comparing coefficients:

\frac{m}{h / a ^{2}} = \frac{- 1}{- k / b ^{2}} = \frac{a / m}{- 1} \Rightarrow m = \frac{h}{k} \cdot \frac{b ^{2}}{a ^{2}} and m = - \frac{ak}{b ^{2}}

Equating the two values of $m$ : $\frac{h b ^{2}}{k a ^{2}} = - \frac{ak}{b ^{2}}$ , so $k^{2} = - \frac{b ^{4}}{a ^{3}} h$ .

The locus of $P$ is $y^{2} = - \frac{b ^{4}}{a ^{3}} x$ .

8. Chord with a Given Middle Point JEE Advanced level

The equation of the chord of the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ whose middle point is $(x_{1}, y_{1})$ is

T = S_{1}, that is, \frac{x x _{1}}{a ^{2}} - \frac{y y _{1}}{b ^{2}} - 1 = \frac{x _{1}^{2}}{a ^{2}} - \frac{y _{1}^{2}}{b ^{2}} - 1

Solved Example 8

Find the locus of the midpoint of focal chords of the hyperbola

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1

Solution:

Let $P \equiv (h, k)$ be the midpoint. The chord with this midpoint is

\frac{x h}{a ^{2}} - \frac{y k}{b ^{2}} - 1 = \frac{h ^{2}}{a ^{2}} - \frac{k ^{2}}{b ^{2}} - 1

Since it is a focal chord, it passes through a focus, either $(a e, 0)$ or $(- a e, 0)$ .

If it passes through $(a e, 0)$ : $\frac{e h}{a} = \frac{h ^{2}}{a ^{2}} - \frac{k ^{2}}{b ^{2}}$ , so the locus is $\frac{e x}{a} = \frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}}$ .

If it passes through $(- a e, 0)$ , the locus is $- \frac{e x}{a} = \frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}}$ .

Solved Example 9

Find the condition on

a

and

b

for which two distinct chords of the hyperbola

\frac{x ^{2}}{2 a ^{2}} - \frac{y ^{2}}{2 b ^{2}} = 1

passing through

(a, b)

are bisected by the line

x + y = b

Solution:

Let the line $x + y = b$ bisect the chord at $P (α, b - α)$ . The chord with this midpoint is

\frac{xα}{2 a ^{2}} - \frac{y ( b - α )}{2 b ^{2}} = \frac{α ^{2}}{2 a ^{2}} - \frac{( b - α ) ^{2}}{2 b ^{2}}

Since it passes through $(a, b)$ :

\frac{α}{2 a} - \frac{b - α}{2 b} = \frac{α ^{2}}{2 a ^{2}} - \frac{( b - α ) ^{2}}{2 b ^{2}} α^{2} (\frac{1}{a ^{2}} - \frac{1}{b ^{2}}) + α (\frac{1}{b} - \frac{1}{a}) = 0

So $α = 0$ or $α = \frac{1}{\frac{1}{a} + \frac{1}{b}}$ . For two distinct values of $α$ , the coefficient $\frac{1}{a ^{2}} - \frac{1}{b ^{2}}$ must be non-zero and $\frac{1}{a} + \frac{1}{b}$ must be non-zero.

Hence the required condition is $a \neq = \pm b$ .

Solved Example 10

Find the locus of the midpoint of the chords of the hyperbola

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1

which subtend a right angle at the origin.

Solution:

Let $(h, k)$ be the midpoint of the chord. Its equation is

\frac{h x}{a ^{2}} - \frac{k y}{b ^{2}} - 1 = \frac{h ^{2}}{a ^{2}} - \frac{k ^{2}}{b ^{2}} - 1 or \frac{h x}{a ^{2}} - \frac{k y}{b ^{2}} = \frac{h ^{2}}{a ^{2}} - \frac{k ^{2}}{b ^{2}} \dots (1)

The lines joining the origin to the points where chord (1) meets the hyperbola are obtained by homogenising the hyperbola with the help of (1):

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = (\frac{\frac{h x}{a ^{2}} - \frac{k y}{b ^{2}}}{\frac{h ^{2}}{a ^{2}} - \frac{k ^{2}}{b ^{2}}})^{2}

\frac{1}{a ^{2}} (\frac{h ^{2}}{a ^{2}} - \frac{k ^{2}}{b ^{2}})^{2} x^{2} - \frac{1}{b ^{2}} (\frac{h ^{2}}{a ^{2}} - \frac{k ^{2}}{b ^{2}})^{2} y^{2} = \frac{h ^{2}}{a ^{4}} x^{2} + \frac{k ^{2}}{b ^{4}} y^{2} - \frac{2 hk}{a ^{2} b ^{2}} x y \dots (2)

The lines in (2) are at right angles if coefficient of $x^{2}$ + coefficient of $y^{2}$ $= 0$ :

\frac{1}{a ^{2}} (\frac{h ^{2}}{a ^{2}} - \frac{k ^{2}}{b ^{2}})^{2} - \frac{h ^{2}}{a ^{4}} - \frac{1}{b ^{2}} (\frac{h ^{2}}{a ^{2}} - \frac{k ^{2}}{b ^{2}})^{2} - \frac{k ^{2}}{b ^{4}} = 0

(\frac{h ^{2}}{a ^{2}} - \frac{k ^{2}}{b ^{2}})^{2} (\frac{1}{a ^{2}} - \frac{1}{b ^{2}}) = \frac{h ^{2}}{a ^{4}} + \frac{k ^{2}}{b ^{4}}

Hence the locus of $(h, k)$ is $(\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}})^{2} (\frac{1}{a ^{2}} - \frac{1}{b ^{2}}) = \frac{x ^{2}}{a ^{4}} + \frac{y ^{2}}{b ^{4}}$ .

9. Diameter of a Hyperbola JEE Advanced level

The locus of the middle points of a system of parallel chords with slope $m$ of a hyperbola is called its diameter. It is a straight line through the centre with equation

y = \frac{b ^{2}}{a ^{2} m} x

Why: the chord with midpoint $(h, k)$ is $T = S_{1}$ , whose slope is $\frac{b ^{2} h}{a ^{2} k}$ . Setting this equal to $m$ gives $k = \frac{b ^{2}}{a ^{2} m} h$ .

Note: All diameters of the hyperbola pass through its centre.

10. Important Properties of Tangents and Normals

The locus of the feet of the perpendiculars drawn from a focus of the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ upon any tangent is its auxiliary circle $x^{2} + y^{2} = a^{2}$ , and the product of these perpendiculars from the two foci is $b^{2}$ .

Figure 5: The feet

T

and

T^{'}

of the perpendiculars from the foci to any tangent lie on the auxiliary circle

x^{2} + y^{2} = a^{2}

, and

S T \cdot S^{'} T^{'} = b^{2}

The portion of a tangent between the point of contact and a directrix subtends a right angle at the corresponding focus.
The tangent and normal at any point of a hyperbola bisect the angles between the focal radii. The tangent bisects the angle $∠ S P S^{'}$ and the normal bisects the exterior angle.
Reflection property: an incoming light ray aimed towards one focus is reflected from the outer surface of the hyperbola towards the other focus. Equivalently, a ray from one focus is reflected as if it came from the other focus.
If an ellipse and a hyperbola have the same foci, they cut at right angles at every common point. For example, the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ and the hyperbola $\frac{x ^{2}}{a ^{2} - k ^{2}} - \frac{y ^{2}}{k ^{2} - b ^{2}} = 1$ , where $a > k > b > 0$ , are confocal and therefore orthogonal.
The foci of the hyperbola and the points $P$ and $Q$ where any tangent meets the tangents at the vertices are concyclic, with $P Q$ as a diameter of the circle.

Figure 6: Reflection property. A ray from

S (5, 0)

meets

9 x^{2} - 16 y^{2} = 144

P (8, 33)

and reflects along the line

S^{'} P

extended. The tangent at

P

bisects

∠ S P S^{'}

Solved Example 11

A ray originating from the point

(5, 0)

is incident on the hyperbola

9 x^{2} - 16 y^{2} = 144

at the point

P

with abscissa 8. Find the equation of the reflected ray after the first reflection, with

P

lying in the first quadrant.

Solution:

The hyperbola is $\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1 \dots (1)$ , so $a = 4$ , $b = 3$ , $e = \frac{5}{4}$ and the foci are $S (5, 0)$ and $S^{'} (- 5, 0)$ . The ray starts from the focus $S$ .

Let $P \equiv (8, α)$ . Since it lies on (1): $\frac{64}{16} - \frac{α ^{2}}{9} = 1$ , so $α^{2} = 27$ and $α = 33$ (as $P$ is in the first quadrant). Hence $P \equiv (8, 33)$ .

By the reflection property, the reflected ray travels along the line through $S^{'} (- 5, 0)$ and $P (8, 33)$ :

y - 33 = \frac{0 - 3 3}{- 5 - 8} (x - 8) \Rightarrow 13 y - 393 = 33 x - 243

The reflected ray is $33 x - 13 y + 153 = 0$ .

11. Practice Problems

Show that the line $x cos α + y sin α = p$ touches the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ if $a^{2} cos^{2} α - b^{2} sin^{2} α = p^{2}$ .
Hint: write the line as $y = - x cot α + p cosec α$ and apply $c^{2} = a^{2} m^{2} - b^{2}$ .
For what value of $λ$ does the line $y = 2 x + λ$ touch the hyperbola $16 x^{2} - 9 y^{2} = 144$ ?
Answer: $λ = \pm 25$
Find the equation of the tangent to the hyperbola $x^{2} - y^{2} = 1$ which is parallel to the line $4 y = 5 x + 7$ .
Answer: $4 y = 5 x \pm 3$
Prove that the line $ℓ x + m y - n = 0$ will be a normal to the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ if $\frac{a ^{2}}{ℓ ^{2}} - \frac{b ^{2}}{m ^{2}} = \frac{( a ^{2} + b ^{2} ) ^{2}}{n ^{2}}$ .
Hint: compare with the normal $a x cos θ + b y cot θ = a^{2} + b^{2}$ and eliminate $θ$ .
Find the locus of the foot of the perpendicular from the centre upon any normal to the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ .
Answer: $(x^{2} + y^{2})^{2} (a^{2} y^{2} - b^{2} x^{2}) = x^{2} y^{2} (a^{2} + b^{2})^{2}$
Find the equation of the chord of the hyperbola $\frac{x ^{2}}{36} - \frac{y ^{2}}{9} = 1$ which is bisected at $(2, 1)$ .
Answer: $x = 2 y$
Find the point $P$ from which the pair of tangents $P A$ and $P B$ are drawn to the hyperbola $\frac{x ^{2}}{25} - \frac{y ^{2}}{16} = 1$ in such a way that $(5, 2)$ bisects $A B$ .
Answer: $P \equiv (\frac{20}{3}, \frac{8}{3})$
From points on the circle $x^{2} + y^{2} = a^{2}$ , tangents are drawn to the hyperbola $x^{2} - y^{2} = a^{2}$ . Prove that the locus of the middle points of the chords of contact is the curve $(x^{2} - y^{2})^{2} = a^{2} (x^{2} + y^{2})$ .
Hint: the chord of contact from $(p, q)$ must coincide with the chord $T = S_{1}$ of midpoint $(h, k)$ ; then use $p^{2} + q^{2} = a^{2}$ .

Common Mistakes to Avoid

Watch out

Using $y = m x \pm a^{2} m^{2} - b^{2}$ for any slope. A real tangent of slope $m$ exists only when $∣ m ∣ > \frac{b}{a}$ .
Calling a line parallel to an asymptote a tangent just because it meets the hyperbola at one point. The condition for tangency is $c^{2} = a^{2} m^{2} - b^{2}$ , not "one intersection point".
Reporting two tangents from a point on an asymptote. There, $S S_{1} = T^{2}$ splits into the real tangent and the asymptote itself, so only one tangent exists.
Copying the ellipse normal $\frac{a ^{2} x}{x _{1}} - \frac{b ^{2} y}{y _{1}} = a^{2} - b^{2}$ . For the hyperbola both signs change: $\frac{a ^{2} x}{x _{1}} + \frac{b ^{2} y}{y _{1}} = a^{2} + b^{2}$ .
Writing the director circle as $x^{2} + y^{2} = a^{2} + b^{2}$ (that is the ellipse result). For the hyperbola it is $x^{2} + y^{2} = a^{2} - b^{2}$ , and it does not exist when $b > a$ .
Mixing up $T = 0$ (chord of contact from an external point) and $T = S_{1}$ (chord bisected at a given point).
Taking the diameter as $y = - \frac{b ^{2}}{a ^{2} m} x$ . That is the ellipse result; for the hyperbola the sign is positive.

Frequently Asked Questions

What is the equation of a tangent to a hyperbola in slope form?

For $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ , the tangents of slope $m$ are $y = m x \pm a^{2} m^{2} - b^{2}$ . They are real only when $∣ m ∣ > \frac{b}{a}$ , and the point of contact is $(- \frac{a ^{2} m}{c}, - \frac{b ^{2}}{c})$ , where $c$ is the chosen intercept.

What is the condition for a line to touch a hyperbola?

The line $y = m x + c$ touches $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ if and only if $c^{2} = a^{2} m^{2} - b^{2}$ . For a line written as $ℓ x + m y + n = 0$ , the equivalent condition is $a^{2} ℓ^{2} - b^{2} m^{2} = n^{2}$ .

What is the equation of the normal to a hyperbola?

At $(x_{1}, y_{1})$ the normal is $\frac{a ^{2} x}{x _{1}} + \frac{b ^{2} y}{y _{1}} = a^{2} + b^{2}$ . At the parametric point $(a sec θ, b tan θ)$ it is $a x cos θ + b y cot θ = a^{2} + b^{2}$ . In slope form it is $y = m x \pm \frac{( a ^{2} + b ^{2} ) m}{a ^{2} - b ^{2} m ^{2}}$ .

What is the director circle of a hyperbola?

It is the locus of points from which two perpendicular tangents can be drawn: $x^{2} + y^{2} = a^{2} - b^{2}$ . It is a real circle only when $a > b$ . For a rectangular hyperbola it shrinks to the centre, and when $b > a$ no perpendicular tangents exist at all.

What is the difference between chord of contact and chord with a given midpoint?

The chord of contact joins the points where the two tangents from an external point $(x_{1}, y_{1})$ touch the curve; its equation is $T = 0$ . The chord bisected at $(x_{1}, y_{1})$ has that point as its midpoint; its equation is $T = S_{1}$ . Both lines have the same slope but different intercepts.

How many tangents can be drawn from a point to a hyperbola?

Two tangents from a point outside the curve that is not on an asymptote, one tangent from a point on the curve, one from a point on an asymptote other than the centre, and none from the centre or from inside. The asymptote case is a common JEE Advanced trap because $S S_{1} = T^{2}$ still gives two lines.

What is the reflection property of a hyperbola?

The tangent at any point $P$ bisects the angle between the focal radii $S P$ and $S^{'} P$ . So a ray from one focus is reflected as if it came from the other focus, and a ray aimed at one focus from outside is reflected towards the other focus. This is why confocal ellipses and hyperbolas cut at right angles.

Are tangents and normals of a hyperbola asked in JEE Main and JEE Advanced?

Yes. JEE Advanced explicitly lists equations of tangent and normal to conics along with locus problems, and JEE Main papers regularly include tangent condition and normal questions on the hyperbola. Pair of tangents, chord of contact and director circle appear mainly in JEE Advanced-level problems.

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