Asymptotes And Rectangular Hyperbola

Q: Does a hyperbola ever intersect its asymptotes?

No. A point common to x^2/a^2 - y^2/b^2 = 1 and x^2/a^2 - y^2/b^2 = 0 would need 1 = 0, which is impossible. The curve gets arbitrarily close to its asymptotes but never meets them at a finite point.

Q: How do you find the asymptotes of a hyperbola given by a general equation?

Replace the constant term of the equation by lambda and choose lambda so that the new equation represents a pair of straight lines, using abc + 2fgh - af^2 - bg^2 - ch^2 = 0. Factorise the result to get the two asymptotes. As a check, both lines must pass through the centre found by setting both partial derivatives of the equation to zero.

Q: Why is the eccentricity of a rectangular hyperbola root 2?

For any hyperbola e^2 = 1 + b^2/a^2. In a rectangular hyperbola a = b, so e^2 = 2 and e = sqrt(2). The same value holds for xy = c^2, because rotating the axes does not change the shape of the curve.

Q: What are the parametric form, tangent and normal of a rectangular hyperbola referred to its asymptotes?

The parametric point is (ct, c/t) with t not equal to 0. The tangent at t is x/t + ty = 2c, the normal at t is xt^3 - yt = c(t^4 - 1), and the chord joining t1 and t2 is x + t1 t2 y = c(t1 + t2).

MathsHyperbolaFor JEE aspirants

The asymptotes of the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ are the lines $\frac{x}{a} \pm \frac{y}{b} = 0$ , which the curve approaches more and more closely but never meets at a finite point. A rectangular hyperbola has equal axes $(a = b)$ , perpendicular asymptotes and eccentricity $2$ ; referred to its asymptotes it becomes $x y = c^{2}$ . This page explains asymptotes, their properties, finding the asymptotes of a general hyperbola, and every standard result for the rectangular hyperbola $x y = c^{2}$ , with solved examples for JEE Main and JEE Advanced.

Key Formulas - Quick Reference

Asymptotes of $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ : $\frac{x}{a} \pm \frac{y}{b} = 0$ ; pair of asymptotes $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 0$
Hyperbola, asymptotes and conjugate differ only in the constant: $= 1$ , $= 0$ , $= - 1$
If the angle between the asymptotes is $2 θ$ , then $e = sec θ$
$a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ is a hyperbola if $h^{2} > ab$ and $Δ \neq = 0$ ; rectangular if also $a + b = 0$
Tangent at $P$ cuts off $△ C QR$ of area $ab$ from the asymptotes, with $P$ the midpoint of $QR$
Rectangular hyperbola $x^{2} - y^{2} = a^{2}$ : asymptotes $y = \pm x$ , $e = 2$
$x y = c^{2}$ : vertices $(c, c), (- c, - c)$ ; foci $(2 c, 2 c), (- 2 c, - 2 c)$ ; directrices $x + y = \pm 2 c$
$x y = c^{2}$ : point $(c t, \frac{c}{t})$ ; tangent $\frac{x}{t} + t y = 2 c$ ; normal $x t^{3} - y t = c (t^{4} - 1)$
$x y = c^{2}$ : chord $x + t_{1} t_{2} y = c (t_{1} + t_{2})$ ; chord with midpoint $(h, k)$ : $k x + h y = 2 hk$

1. Asymptotes of a Hyperbola

Definition: If the length of the perpendicular let fall from a point on a hyperbola to a straight line tends to zero as the point on the hyperbola moves to infinity along the hyperbola, then the straight line is called an asymptote of the hyperbola.

Equations of the asymptotes: $\frac{x}{a} + \frac{y}{b} = 0$ and $\frac{x}{a} - \frac{y}{b} = 0$
Pair of asymptotes: $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 0$

Figure 1: The asymptotes

\frac{x}{a} \pm \frac{y}{b} = 0

are the extended diagonals of the rectangle through

x = \pm a

and

y = \pm b

. The angle

2 θ

between them satisfies

tan θ = \frac{b}{a}

Why these lines are the asymptotes

Substituting $y = m x + c$ into the hyperbola gives $(b^{2} - a^{2} m^{2}) x^{2} - 2 a^{2} m c x - a^{2} (c^{2} + b^{2}) = 0$ . A line is an asymptote when both roots of this quadratic are infinite, which happens when the coefficients of $x^{2}$ and $x$ are both zero:

b^{2} - a^{2} m^{2} = 0 and a^{2} m c = 0 \Rightarrow m = \pm \frac{b}{a}, c = 0

So the asymptotes are $y = \pm \frac{b}{a} x$ , that is, $\frac{x}{a} \pm \frac{y}{b} = 0$ .

Note:

(i) A hyperbola and its conjugate have the same asymptotes.
(ii) The equation of the pair of asymptotes differs from the equation of the hyperbola (or the conjugate hyperbola) by the constant term only.
(iii) The asymptotes pass through the centre of the hyperbola and are equally inclined to the transverse axis. Hence the bisectors of the angles between the asymptotes are the principal axes of the hyperbola.
(iv) The asymptotes of a hyperbola are the diagonals of the rectangle formed by the lines drawn through the extremities of each axis parallel to the other axis.
(v) To find the centre of a hyperbola given by a general second-degree equation $f (x, y) = 0$ , solve $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$ together. Their point of intersection is the centre.

Remarks:

(i) No tangent to the hyperbola can be drawn from its centre.
(ii) Only one tangent to the hyperbola can be drawn from a point lying on an asymptote, other than the centre.
(iii) Two tangents can be drawn to the hyperbola from any external point that does not lie on an asymptote.

2. Asymptotes of a General Hyperbola JEE Advanced level

For a general second-degree equation $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ (here $a, b, c, f, g, h$ are coefficients, not semi-axes), define

Δ = ab c + 2 f g h - a f^{2} - b g^{2} - c h^{2}

The equation represents a hyperbola if $h^{2} > ab$ and $Δ \neq = 0$ .
It is a rectangular hyperbola if, in addition, $a + b = 0$ .
If $Δ = 0$ , it represents a pair of straight lines.

Method: since the hyperbola and its asymptotes differ only in the constant term, write the pair of asymptotes as the given equation with the constant changed to $λ$ . Choose $λ$ so that the new equation represents two straight lines, that is, $Δ = 0$ .

Solved Example 1

Find the asymptotes of

x y - 3 y - 2 x = 0

Solution:

Here $h^{2} = \frac{1}{4} > ab = 0$ , so the curve is a hyperbola. Since a hyperbola and its asymptotes differ in the constant term only, the pair of asymptotes is

x y - 3 y - 2 x + λ = 0 \dots (1)

where $λ$ is chosen so that (1) represents two straight lines. With $a = 0, b = 0, h = \frac{1}{2}, g = - 1, f = - \frac{3}{2}, c = λ$ , the condition $ab c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0$ gives

0 + 2 \times (- \frac{3}{2}) \times (- 1) \times \frac{1}{2} - 0 - 0 - λ (\frac{1}{2})^{2} = 0 \Rightarrow \frac{3}{2} - \frac{λ}{4} = 0 \Rightarrow λ = 6

From (1), the pair of asymptotes is $x y - 3 y - 2 x + 6 = 0$ , that is, $(y - 2) (x - 3) = 0$ .

Check using note (v): $\frac{\partial f}{\partial x} = y - 2 = 0$ and $\frac{\partial f}{\partial y} = x - 3 = 0$ give the centre $(3, 2)$ , which lies on both lines.

The asymptotes are $x - 3 = 0$ and $y - 2 = 0$ .

Solved Example 2

The asymptotes of a hyperbola having centre at the point

(1, 2)

are parallel to the lines

2 x + 3 y = 0

and

3 x + 2 y = 0

. If the hyperbola passes through the point

(5, 3)

, show that its equation is

(2 x + 3 y - 8) (3 x + 2 y - 7) = 154

Solution:

Let the asymptotes be $2 x + 3 y + λ = 0$ and $3 x + 2 y + μ = 0$ . Since the asymptotes pass through the centre $(1, 2)$ : $2 + 6 + λ = 0$ gives $λ = - 8$ , and $3 + 4 + μ = 0$ gives $μ = - 7$ .

So the asymptotes are $2 x + 3 y - 8 = 0$ and $3 x + 2 y - 7 = 0$ . Since the hyperbola differs from the pair of asymptotes only in the constant term, let its equation be

(2 x + 3 y - 8) (3 x + 2 y - 7) + v = 0 \dots (1)

It passes through $(5, 3)$ :

(10 + 9 - 8) (15 + 6 - 7) + v = 0 \Rightarrow 11 \times 14 + v = 0 \Rightarrow v = - 154

Putting this value in (1), the required hyperbola is $(2 x + 3 y - 8) (3 x + 2 y - 7) = 154$ .

3. Important Properties of Asymptotes JEE Advanced level

The tangent at any point $P$ on $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ with centre $C$ meets the asymptotes in $Q$ and $R$ and cuts off a $△ C QR$ of constant area $ab$ . The portion $QR$ of the tangent intercepted between the asymptotes is bisected at the point of contact $P$ .
Consequently, the locus of the centre of the circle circumscribing $△ C QR$ is the curve $4 (a^{2} x^{2} - b^{2} y^{2}) = (a^{2} + b^{2})^{2}$ . For a rectangular hyperbola this locus is the hyperbola itself.

Figure 2: The tangent at

P

meets the asymptotes at

Q

and

R

P

is the midpoint of

QR

, and the area of

△ C QR

is always

ab

If from any point on an asymptote a straight line is drawn perpendicular to the transverse axis, the product of the segments of this line intercepted between the point and the curve is always equal to $b^{2}$ , the square of the semi-conjugate axis.
The perpendiculars from the foci on either asymptote meet it at the same points as the corresponding directrix, and these common points lie on the auxiliary circle. For the focus $(a e, 0)$ and asymptote $y = \frac{b}{a} x$ , the foot is $(\frac{a}{e}, \frac{b}{e})$ .
If the angle between the asymptotes of $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ is $2 θ$ , then the eccentricity of the hyperbola is $sec θ$ . This follows from $tan θ = \frac{b}{a}$ , so $sec^{2} θ = 1 + \frac{b ^{2}}{a ^{2}} = e^{2}$ .

4. Rectangular Hyperbola (Equilateral Hyperbola)

The particular kind of hyperbola in which the lengths of the transverse and conjugate axes are equal is called an equilateral or rectangular hyperbola.

Since $a = b$ , the equation becomes

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

whose asymptotes are $y = \pm x$ . They are perpendicular, which is why the curve is called rectangular. Its eccentricity is

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

Figure 3: Rectangular hyperbola

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

. Its asymptotes

y = \pm x

are perpendicular, each at

4 5^{\circ}

to the transverse axis.

Rotating this system through an angle of $4 5^{\circ}$ in the clockwise direction gives another form of the equation of a rectangular hyperbola:

x y = c^{2}, where c^{2} = \frac{a ^{2}}{2}

5. Rectangular Hyperbola $x y = c^{2}$ JEE Advanced level

This form is referred to its asymptotes as the axes of coordinates. Its standard results are:

Property	For $x y = c^{2}$
Centre and asymptotes	Centre $(0, 0)$ ; asymptotes $x = 0$ and $y = 0$
Transverse and conjugate axes	Along $y = x$ and $y = - x$
Vertices	$(c, c)$ and $(- c, - c)$
Foci	$(2 c, 2 c)$ and $(- 2 c, - 2 c)$
Directrices	$x + y = \pm 2 c$
Eccentricity	$2$
Latus rectum $(ℓ)$	$ℓ = 22 c =$ T.A. $=$ C.A.
Parametric equations	$x = c t, y = \frac{c}{t}$ , where $t \in R - {0}$
Chord joining $P (t_{1})$ and $Q (t_{2})$	$x + t_{1} t_{2} y = c (t_{1} + t_{2})$
Tangent at $P (x_{1}, y_{1})$	$\frac{x}{x _{1}} + \frac{y}{y _{1}} = 2$
Tangent at $P (t)$	$\frac{x}{t} + t y = 2 c$
Normal at $P (t)$	$x t^{3} - y t = c (t^{4} - 1)$
Chord with middle point $(h, k)$	$k x + h y = 2 hk$

Figure 4: Rectangular hyperbola

x y = c^{2}

with vertices

(c, c)

and

(- c, - c)

, foci

(2 c, 2 c)

and

(- 2 c, - 2 c)

, and directrices

x + y = \pm 2 c

Solved Example 3

A triangle has its vertices on a rectangular hyperbola. Prove that the orthocentre of the triangle also lies on the same hyperbola.

Solution:

Let $t_{1}, t_{2}$ and $t_{3}$ be the parameters of the vertices of $△ A B C$ on the rectangular hyperbola $x y = c^{2}$ . The coordinates of $A$ , $B$ and $C$ are $(c t_{1}, \frac{c}{t _{1}})$ , $(c t_{2}, \frac{c}{t _{2}})$ and $(c t_{3}, \frac{c}{t _{3}})$ respectively.

Slope of $B C$ :

\frac{c ( \frac{1}{t _{3}} - \frac{1}{t _{2}} )}{c ( t _{3} - t _{2} )} = - \frac{1}{t _{2} t _{3}}

So the slope of the altitude $A D$ is $t_{2} t_{3}$ , and its equation is

y - \frac{c}{t _{1}} = t_{2} t_{3} (x - c t_{1}) or t_{1} y - c = x t_{1} t_{2} t_{3} - c t_{1}^{2} t_{2} t_{3} \dots (1)

Similarly, the altitude $B E$ is

t_{2} y - c = x t_{1} t_{2} t_{3} - c t_{1} t_{2}^{2} t_{3} \dots (2)

Subtracting (2) from (1): $(t_{1} - t_{2}) y = - c t_{1} t_{2} t_{3} (t_{1} - t_{2})$ , so $y = - c t_{1} t_{2} t_{3}$ . Then $x = - \frac{c}{t _{1} t _{2} t _{3}}$ . The orthocentre is

H (- \frac{c}{t _{1} t _{2} t _{3}}, - c t_{1} t_{2} t_{3})

Since $x \cdot y = c^{2}$ , the orthocentre lies on $x y = c^{2}$ .

Figure 5: For

△ A B C

inscribed in

x y = c^{2}

, the altitudes meet at

H (- \frac{c}{t _{1} t _{2} t _{3}}, - c t_{1} t_{2} t_{3})

, which also lies on the hyperbola.

Solved Example 4

A

B

C

are three points on the rectangular hyperbola

x y = c^{2}

. Find (i) the area of

△ A B C

and (ii) the area of the triangle formed by the tangents at

A

B

and

C

Solution:

Let the coordinates of $A$ , $B$ and $C$ be $(c t_{1}, \frac{c}{t _{1}})$ , $(c t_{2}, \frac{c}{t _{2}})$ and $(c t_{3}, \frac{c}{t _{3}})$ .

(i) Area of $△ A B C$ :

Δ_{1} = \frac{1}{2} c t_{1} \frac{c}{t _{1}} c t_{2} \frac{c}{t _{2}} + c t_{2} \frac{c}{t _{2}} c t_{3} \frac{c}{t _{3}} + c t_{3} \frac{c}{t _{3}} c t_{1} \frac{c}{t _{1}}

= \frac{c ^{2}}{2} \frac{t _{1}}{t _{2}} - \frac{t _{2}}{t _{1}} + \frac{t _{2}}{t _{3}} - \frac{t _{3}}{t _{2}} + \frac{t _{3}}{t _{1}} - \frac{t _{1}}{t _{3}} = \frac{c ^{2}}{2∣ t _{1} t _{2} t _{3} ∣} t_{1}^{2} t_{3} - t_{2}^{2} t_{3} + t_{1} t_{2}^{2} - t_{1} t_{3}^{2} + t_{2} t_{3}^{2} - t_{1}^{2} t_{2} = \frac{c ^{2}}{2∣ t _{1} t _{2} t _{3} ∣} ∣ (t_{1} - t_{2}) (t_{2} - t_{3}) (t_{3} - t_{1}) ∣

(ii) The tangents at $A$ , $B$ and $C$ are

x + t_{1}^{2} y - 2 c t_{1} = 0, x + t_{2}^{2} y - 2 c t_{2} = 0, x + t_{3}^{2} y - 2 c t_{3} = 0

The area of the triangle formed by three lines is

Δ_{2} = \frac{1}{2∣ C _{1} C _{2} C _{3} ∣} 1 t_{1}^{2} - 2 c t_{1} 1 t_{2}^{2} - 2 c t_{2} 1 t_{3}^{2} - 2 c t_{3}^{2} \dots (1)

where $C_{1}, C_{2}, C_{3}$ are the cofactors of $- 2 c t_{1}, - 2 c t_{2}, - 2 c t_{3}$ :

C_{1} = 1 t_{2}^{2} 1 t_{3}^{2} = t_{3}^{2} - t_{2}^{2}, C_{2} = - 1 t_{1}^{2} 1 t_{3}^{2} = t_{1}^{2} - t_{3}^{2}, C_{3} = 1 t_{1}^{2} 1 t_{2}^{2} = t_{2}^{2} - t_{1}^{2}

The determinant in (1) equals $- 2 c (t_{1} - t_{2}) (t_{2} - t_{3}) (t_{3} - t_{1})$ up to sign, so from (1):

Δ_{2} = \frac{4 c ^{2} ( t _{1} - t _{2} ) ^{2} ( t _{2} - t _{3} ) ^{2} ( t _{3} - t _{1} ) ^{2}}{2 ∣ ( t _{3}^{2} - t _{2}^{2} ) ( t _{1}^{2} - t _{3}^{2} ) ( t _{2}^{2} - t _{1}^{2} ) ∣} = 2 c^{2} \frac{( t _{1} - t _{2} ) ( t _{2} - t _{3} ) ( t _{3} - t _{1} )}{( t _{1} + t _{2} ) ( t _{2} + t _{3} ) ( t _{3} + t _{1} )}

So the areas are (i) $\frac{c ^{2}}{2∣ t _{1} t _{2} t _{3} ∣} ∣ (t_{1} - t_{2}) (t_{2} - t_{3}) (t_{3} - t_{1}) ∣$ and (ii) $2 c^{2} \frac{( t _{1} - t _{2} ) ( t _{2} - t _{3} ) ( t _{3} - t _{1} )}{( t _{1} + t _{2} ) ( t _{2} + t _{3} ) ( t _{3} + t _{1} )}$ .

Solved Example 5

Prove that perpendicular focal chords of a rectangular hyperbola are equal.

Solution:

Let the rectangular hyperbola be $x^{2} - y^{2} = a^{2}$ , and let the focal chords $P Q$ and $D E$ through the focus $S (a 2, 0)$ be

y = m x + c \dots (1) and y = m_{1} x + c_{1} \dots (2)

Since $P Q ⊥ D E$ , $m m_{1} = - 1 \dots (3)$ . As $P Q$ passes through $S (a 2, 0)$ , from (1): $0 = ma 2 + c$ , so $c^{2} = 2 a^{2} m^{2} \dots (4)$ .

Figure 6: Perpendicular focal chords

P Q

and

D E

through

S (a 2, 0)

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

are equal in length.

Let $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ be the coordinates of $P$ and $Q$ . Since both lie on (1), $y_{1} - y_{2} = m (x_{1} - x_{2})$ , so

(P Q)^{2} = (x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2} = (x_{1} - x_{2})^{2} (1 + m^{2}) \dots (5)

Solving $y = m x + c$ with $x^{2} - y^{2} = a^{2}$ : $x^{2} - (m x + c)^{2} = a^{2}$ , or $(m^{2} - 1) x^{2} + 2 m c x + (a^{2} + c^{2}) = 0$ . So

x_{1} + x_{2} = - \frac{2 m c}{m ^{2} - 1}, x_{1} x_{2} = \frac{a ^{2} + c ^{2}}{m ^{2} - 1}

(x_{1} - x_{2})^{2} = (x_{1} + x_{2})^{2} - 4 x_{1} x_{2} = \frac{4 m ^{2} c ^{2}}{( m ^{2} - 1 ) ^{2}} - \frac{4 ( a ^{2} + c ^{2} )}{m ^{2} - 1} = \frac{4 { c ^{2} - a ^{2} m ^{2} + a ^{2} }}{( m ^{2} - 1 ) ^{2}} = \frac{4 a ^{2} ( m ^{2} + 1 )}{( m ^{2} - 1 ) ^{2}} {∵ c^{2} = 2 a^{2} m^{2}}

From (5):

(P Q)^{2} = 4 a^{2} (\frac{m ^{2} + 1}{m ^{2} - 1})^{2}

Similarly, using (3),

(D E)^{2} = 4 a^{2} (\frac{m _{1}^{2} + 1}{m _{1}^{2} - 1})^{2} = 4 a^{2} (\frac{\frac{1}{m ^{2}} + 1}{\frac{1}{m ^{2}} - 1})^{2} = 4 a^{2} (\frac{1 + m ^{2}}{1 - m ^{2}})^{2} = (P Q)^{2}

Thus $(P Q)^{2} = (D E)^{2}$ , so $P Q = D E$ . Hence perpendicular focal chords of a rectangular hyperbola are equal.

6. Important Properties of the Rectangular Hyperbola JEE Advanced level

A rectangular hyperbola circumscribing a triangle also passes through the orthocentre of that triangle. If $(c t_{i}, \frac{c}{t _{i}})$ , $i = 1, 2, 3$ , are the vertices, the orthocentre is $(- \frac{c}{t _{1} t _{2} t _{3}}, - c t_{1} t_{2} t_{3})$ .
If a circle and the rectangular hyperbola $x y = c^{2}$ meet in the four points $t_{1}, t_{2}, t_{3}$ and $t_{4}$ , then:
(a) $t_{1} t_{2} t_{3} t_{4} = 1$
(b) the centre of the mean position of the four points bisects the distance between the centres of the two curves.
(c) the centre of the circle through the points $t_{1}, t_{2}$ and $t_{3}$ is $(\frac{c}{2} (t_{1} + t_{2} + t_{3} + \frac{1}{t _{1} t _{2} t _{3}}), \frac{c}{2} (\frac{1}{t _{1}} + \frac{1}{t _{2}} + \frac{1}{t _{3}} + t_{1} t_{2} t_{3}))$
Perpendicular focal chords of a rectangular hyperbola are equal (Solved Example 5).
The locus of the circumcentre of the triangle cut off from the asymptotes by any tangent is the rectangular hyperbola itself.

7. Practice Problem

Show that the tangent at any point of a hyperbola cuts off a triangle of constant area from the asymptotes, and that the portion of it intercepted between the asymptotes is bisected at the point of contact.
Hint: the tangent at $(a sec θ, b tan θ)$ meets the asymptotes at $Q (a (sec θ + tan θ), b (sec θ + tan θ))$ and $R (a (sec θ - tan θ), - b (sec θ - tan θ))$ . Find the midpoint of $QR$ and the area of $△ C QR$ , using $sec^{2} θ - tan^{2} θ = 1$ .

Common Mistakes to Avoid

Watch out

Treating an asymptote as a tangent. The hyperbola never meets its asymptotes at a finite point; they only touch "at infinity".
Getting the asymptotes of a general hyperbola by simply deleting the constant term. The constant must be replaced by the value of $λ$ that makes $Δ = 0$ . In Solved Example 1 the constant 0 becomes 6.
Writing the vertices of $x y = c^{2}$ as $(c, 0)$ and $(- c, 0)$ . The transverse axis is $y = x$ , so the vertices are $(c, c)$ and $(- c, - c)$ .
Placing the foci of $x y = c^{2}$ at $(2 c, 0)$ . They lie on $y = x$ at $(2 c, 2 c)$ and $(- 2 c, - 2 c)$ , a distance $2 c$ from the centre.
Sign slips in the normal to $x y = c^{2}$ at $t$ . The correct form is $x t^{3} - y t = c (t^{4} - 1)$ .
Thinking the eccentricity of $x y = c^{2}$ depends on $c$ . Every rectangular hyperbola has $e = 2$ .
Using $tan θ = \frac{a}{b}$ for the half-angle between the asymptotes. For $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ it is $tan θ = \frac{b}{a}$ , measured from the transverse axis, and then $e = sec θ$ .

Frequently Asked Questions

What are the asymptotes of a hyperbola?

Asymptotes are straight lines that the hyperbola approaches as a point on it moves to infinity, with the perpendicular distance tending to zero. For $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ they are $y = \pm \frac{b}{a} x$ . Their joint equation is $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 0$ .

Does a hyperbola ever intersect its asymptotes?

No. A point common to $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ and $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 0$ would need $1 = 0$ , which is impossible. The curve gets arbitrarily close to its asymptotes but never meets them at a finite point.

How do you find the asymptotes of a hyperbola given by a general equation?

Replace the constant term of the equation by $λ$ and choose $λ$ so that the new equation represents a pair of straight lines, using $ab c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0$ . Factorise the result to get the two asymptotes. As a check, both lines must pass through the centre found from $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$ .

What is a rectangular hyperbola?

A rectangular (equilateral) hyperbola has equal transverse and conjugate axes, so $a = b$ and its equation is $x^{2} - y^{2} = a^{2}$ . Its asymptotes $y = \pm x$ are perpendicular. Rotating the axes by $4 5^{\circ}$ onto the asymptotes gives the form $x y = c^{2}$ with $c^{2} = \frac{a ^{2}}{2}$ .

Why is the eccentricity of a rectangular hyperbola root 2?

For any hyperbola $e^{2} = 1 + \frac{b ^{2}}{a ^{2}}$ . In a rectangular hyperbola $a = b$ , so $e^{2} = 2$ and $e = 2$ . The same value holds for $x y = c^{2}$ , because rotating the axes does not change the shape of the curve.

What are the parametric form, tangent and normal of a rectangular hyperbola referred to its asymptotes?

The parametric point is $(c t, \frac{c}{t})$ with $t \neq = 0$ . The tangent at $t$ is $\frac{x}{t} + t y = 2 c$ , the normal at $t$ is $x t^{3} - y t = c (t^{4} - 1)$ , and the chord joining $t_{1}$ and $t_{2}$ is $x + t_{1} t_{2} y = c (t_{1} + t_{2})$ .

How is the angle between the asymptotes related to eccentricity?

If the angle between the asymptotes containing the hyperbola is $2 θ$ , then $tan θ = \frac{b}{a}$ and $e = sec θ$ . For example, perpendicular asymptotes give $θ = 4 5^{\circ}$ and $e = 2$ , which is the rectangular hyperbola.

Are asymptotes and the rectangular hyperbola asked in JEE Main and JEE Advanced?

They are not named separately in the syllabus text, but they come under hyperbola in standard form, tangents and locus problems. JEE Main questions often use the eccentricity $2$ and the asymptote equations, while JEE Advanced problems use $x y = c^{2}$ parametric results such as the orthocentre property and the tangent-triangle area $ab$ .

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1. Asymptotes of a Hyperbola

Why these lines are the asymptotes

2. Asymptotes of a General Hyperbola JEE Advanced level

3. Important Properties of Asymptotes JEE Advanced level

4. Rectangular Hyperbola (Equilateral Hyperbola)

5. Rectangular Hyperbola xy=c2 JEE Advanced level

6. Important Properties of the Rectangular Hyperbola JEE Advanced level

7. Practice Problem

Common Mistakes to Avoid

Frequently Asked Questions

What are the asymptotes of a hyperbola?

Does a hyperbola ever intersect its asymptotes?

How do you find the asymptotes of a hyperbola given by a general equation?

What is a rectangular hyperbola?

Why is the eccentricity of a rectangular hyperbola root 2?

What are the parametric form, tangent and normal of a rectangular hyperbola referred to its asymptotes?

How is the angle between the asymptotes related to eccentricity?

Are asymptotes and the rectangular hyperbola asked in JEE Main and JEE Advanced?

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5. Rectangular Hyperbola $x y = c^{2}$ JEE Advanced level