Introduction to Hyperbola

Q: How do you find the eccentricity of a hyperbola?

For x^2/a^2 - y^2/b^2 = 1, use e = sqrt(1 + b^2/a^2), where a is the semi-transverse axis. For the conjugate form y^2/b^2 - x^2/a^2 = 1, use e = sqrt(1 + a^2/b^2). The value always exceeds 1, and it equals sqrt(2) when a = b.

Q: What is a conjugate hyperbola?

The conjugate of x^2/a^2 - y^2/b^2 = 1 is y^2/b^2 - x^2/a^2 = 1, whose transverse and conjugate axes are swapped. Both share the same centre and asymptotes, their four foci form a square, and their eccentricities satisfy 1/e1^2 + 1/e2^2 = 1.

Q: Why does a hyperbola use sec and tan in its parametric form?

Substituting x = a sec(theta) and y = b tan(theta) into x^2/a^2 - y^2/b^2 gives sec^2(theta) - tan^2(theta) = 1, so every value of theta gives a point on the curve. Geometrically, theta is the angle at the centre to the corresponding point (a cos(theta), a sin(theta)) on the auxiliary circle.

MathsHyperbolaFor JEE aspirants

A hyperbola is the locus of a point that moves in a plane so that the ratio of its distance from a fixed point (the focus) to its distance from a fixed line (the directrix) is a constant $e > 1$ . Its standard equation is $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ , where $b^{2} = a^{2} (e^{2} - 1)$ . This page explains every basic term of the hyperbola, including eccentricity, foci, directrices, axes, vertices and latus rectum, along with the conjugate hyperbola, auxiliary circle, parametric form and position of a point, with solved examples for JEE Main and JEE Advanced.

Key Formulas - Quick Reference

Standard equation: $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ , with $b^{2} = a^{2} (e^{2} - 1)$
Eccentricity: $e^{2} = 1 + \frac{b ^{2}}{a ^{2}}$ , always $e > 1$
Foci $(\pm a e, 0)$ ; directrices $x = \pm \frac{a}{e}$ ; vertices $(\pm a, 0)$
Transverse axis $= 2 a$ ; conjugate axis $= 2 b$
Latus rectum $= \frac{2 b ^{2}}{a} = 2 a (e^{2} - 1)$ ; its ends are $(\pm a e, \pm \frac{b ^{2}}{a})$
Focal distances: $S P = ∣ e x_{1} - a ∣$ , $S^{'} P = ∣ e x_{1} + a ∣$ , so $∣ S^{'} P - S P ∣ = 2 a$
Conjugate hyperbola $\frac{y ^{2}}{b ^{2}} - \frac{x ^{2}}{a ^{2}} = 1$ ; if $e_{1}, e_{2}$ are the two eccentricities, $\frac{1}{e _{1}^{2}} + \frac{1}{e _{2}^{2}} = 1$
Auxiliary circle $x^{2} + y^{2} = a^{2}$ ; parametric point $(a sec θ, b tan θ)$
Position of $(x_{1}, y_{1})$ : $S_{1} = \frac{x _{1}^{2}}{a ^{2}} - \frac{y _{1}^{2}}{b ^{2}} - 1$ is $> 0$ (inside), $= 0$ (on), $< 0$ (outside)

1. Definition of a Hyperbola

Hyperbolic curves are of special importance in science and technology, especially in astronomy and space studies. For example, a body that passes the Sun only once, without being captured into orbit, moves along a hyperbolic path. In this chapter we study the characteristics of such curves.

Definition: A hyperbola is the locus of a point moving in a plane such that the ratio of its distance from a fixed point to its distance from a fixed line is a constant greater than 1. The fixed point (focus) does not lie on the fixed line (directrix).

\frac{P S}{P M} = e > 1, e = eccentricity

Figure 1: Focus-directrix definition. For every point

P

on the hyperbola,

\frac{P S}{P M} = e > 1

Hyperbola as a conic section

A hyperbola is one of the conic sections, the curves obtained when a plane cuts a right circular double cone. Let $α$ be the semi-vertical angle of the cone and $β$ the angle between the cutting plane and the axis of the cone. When $β < α$ , the plane cuts both nappes of the cone and the section is a hyperbola.

Figure 2: A plane making angle

β

with the axis of a double cone, with

β < α

, cuts both nappes. The section is a hyperbola.

Angle between plane and axis	Section obtained
$β = 9 0^{\circ}$	Circle
$α < β < 9 0^{\circ}$	Ellipse
$β = α$	Parabola
$0^{\circ} \leq β < α$	Hyperbola

2. Standard Equation of a Hyperbola

Take the focus as $S (a e, 0)$ and the directrix as $x = \frac{a}{e}$ , where $e > 1$ . For any point $P (x, y)$ on the curve, $P M = x - \frac{a}{e}$ , and the definition $P S = e \cdot P M$ gives

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

Standard equation of a hyperbola:

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1, where b^{2} = a^{2} (e^{2} - 1)

Because the equation contains only even powers of $x$ and $y$ , the curve is symmetric about both axes. So $S^{'} (- a e, 0)$ and $x = - \frac{a}{e}$ act as a second focus and a second directrix.

Figure 3: Standard hyperbola

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

with centre

C

, vertices

A, A^{'}

, foci

S, S^{'}

, directrices

x = \pm \frac{a}{e}

, conjugate axis

B^{'} B

and latus rectum

L L^{'}

Basic terms of the hyperbola

Term	Meaning	For $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$
Eccentricity $(e)$	The constant ratio $\frac{P S}{P M}$ , always greater than 1	$e^{2} = 1 + \frac{b ^{2}}{a ^{2}}$
Foci	The two fixed points	$S \equiv (a e, 0)$ and $S^{'} \equiv (- a e, 0)$
Directrices	The fixed lines, one for each focus	$x = \frac{a}{e}$ and $x = - \frac{a}{e}$
Transverse axis	The line segment $A^{'} A$ of length $2 a$ on which both foci $S^{'}$ and $S$ lie	Along the $x$ -axis, length $2 a$
Conjugate axis	The line segment $B^{'} B$ of length $2 b$ between $B^{'} \equiv (0, - b)$ and $B \equiv (0, b)$	Along the $y$ -axis, length $2 b$
Principal axes	The transverse and conjugate axes taken together	The $x$ -axis and $y$ -axis
Vertices	Points where the hyperbola meets its transverse axis	$A \equiv (a, 0)$ and $A^{'} \equiv (- a, 0)$
Centre	The point that bisects every chord of the conic drawn through it	$C \equiv (0, 0)$
Focal chord	A chord that passes through a focus	Any chord through $S$ or $S^{'}$
Double ordinate	A chord perpendicular to the transverse axis	A chord parallel to the $y$ -axis
Latus rectum	A focal chord perpendicular to the transverse axis	Lies on $x = \pm a e$ , length $\frac{2 b ^{2}}{a}$

Length of the latus rectum

Putting $x = a e$ in the equation gives $y^{2} = b^{2} (e^{2} - 1) = \frac{b ^{4}}{a ^{2}}$ , so $y = \pm \frac{b ^{2}}{a}$ . Hence the length $(ℓ)$ of the latus rectum is

ℓ = \frac{2 b ^{2}}{a} = \frac{( C.A. ) ^{2}}{T.A.} = 2 a (e^{2} - 1)

Note:

(i) Length of latus rectum $= 2 e \times$ (distance of the focus from the corresponding directrix). Here the distance is $a e - \frac{a}{e} = \frac{a ( e ^{2} - 1 )}{e}$ .
(ii) The end points of the latera recta are $L \equiv (a e, \frac{b ^{2}}{a})$ , $L^{'} \equiv (a e, - \frac{b ^{2}}{a})$ , $M \equiv (- a e, \frac{b ^{2}}{a})$ and $M^{'} \equiv (- a e, - \frac{b ^{2}}{a})$ .

General note: The fundamental equation of a hyperbola differs from that of an ellipse only in having $- b^{2}$ in place of $b^{2}$ . So many propositions for the hyperbola can be obtained from those for the ellipse simply by changing the sign of $b^{2}$ .

Focal distances of a point

For a point $P (x_{1}, y_{1})$ on the right branch, $P M = x_{1} - \frac{a}{e}$ for the directrix $x = \frac{a}{e}$ . Using $P S = e \cdot P M$ for each focus-directrix pair:

S P = e (x_{1} - \frac{a}{e}) = e x_{1} - a, S^{'} P = e (x_{1} + \frac{a}{e}) = e x_{1} + a

So $S^{'} P - S P = 2 a$ . On the left branch the roles swap and $S P - S^{'} P = 2 a$ . In both cases

∣ S^{'} P - S P ∣ = 2 a = length of the transverse axis

This gives an equivalent definition: a hyperbola is the locus of a point whose distances from two fixed points (the foci) differ by a constant.

Solved Example 1

Find the equation of the hyperbola whose directrix is

2 x + y = 1

, focus is

(1, 2)

and eccentricity is

3

Solution:

Let $P (x, y)$ be any point on the hyperbola and draw $P M$ perpendicular from $P$ to the directrix. By definition, $S P = e \cdot P M$ , so $(S P)^{2} = e^{2} (P M)^{2}$ .

(x - 1)^{2} + (y - 2)^{2} = 3 (\frac{2 x + y - 1}{4 + 1})^{2} 5 (x^{2} + y^{2} - 2 x - 4 y + 5) = 3 (4 x^{2} + y^{2} + 1 + 4 x y - 2 y - 4 x) 7 x^{2} - 2 y^{2} + 12 x y - 2 x + 14 y - 22 = 0

This is the required hyperbola: $7 x^{2} - 2 y^{2} + 12 x y - 2 x + 14 y - 22 = 0$ .

Solved Example 2

Find the eccentricity of the hyperbola whose latus rectum is half of its transverse axis.

Solution:

Let the hyperbola be $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ . Then transverse axis $= 2 a$ and latus rectum $= \frac{2 b ^{2}}{a}$ . According to the question,

\frac{2 b ^{2}}{a} = \frac{1}{2} (2 a) \Rightarrow 2 b^{2} = a^{2} 2 a^{2} (e^{2} - 1) = a^{2} (∵ b^{2} = a^{2} (e^{2} - 1)) 2 e^{2} - 2 = 1 \Rightarrow e^{2} = \frac{3}{2}

Hence the required eccentricity is $e = \frac{3}{2}$ .

3. Hyperbola with Centre at $(h, k)$

When the centre is shifted to $(h, k)$ and the axes stay parallel to the coordinate axes, replace $x$ by $x - h$ and $y$ by $y - k$ . All the results of the standard form carry over after this shift.

Property	$\frac{( x - h ) ^{2}}{a ^{2}} - \frac{( y - k ) ^{2}}{b ^{2}} = 1$	$\frac{( y - k ) ^{2}}{b ^{2}} - \frac{( x - h ) ^{2}}{a ^{2}} = 1$
Centre	$(h, k)$	$(h, k)$
Transverse axis	Along $y = k$ , length $2 a$	Along $x = h$ , length $2 b$
Vertices	$(h \pm a, k)$	$(h, k \pm b)$
Foci	$(h \pm a e, k)$	$(h, k \pm b e)$
Directrices	$x = h \pm \frac{a}{e}$	$y = k \pm \frac{b}{e}$
Eccentricity	$e^{2} = 1 + \frac{b ^{2}}{a ^{2}}$	$e^{2} = 1 + \frac{a ^{2}}{b ^{2}}$
Latus rectum	$\frac{2 b ^{2}}{a}$	$\frac{2 a ^{2}}{b}$

4. Conjugate Hyperbola

Two hyperbolas such that the transverse and conjugate axes of one are respectively the conjugate and transverse axes of the other are called conjugate hyperbolas of each other.

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1 and - \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1 are conjugate hyperbolas of each other.

Figure 4: Conjugate hyperbola

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

. The transverse axis lies along the

y

-axis, foci are

(0, \pm b e)

and directrices are

y = \pm \frac{b}{e}

For the conjugate hyperbola $\frac{y ^{2}}{b ^{2}} - \frac{x ^{2}}{a ^{2}} = 1$ the transverse axis lies along the $y$ -axis, so the roles of $a$ and $b$ are exchanged:

Property	Hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$	Conjugate $\frac{y ^{2}}{b ^{2}} - \frac{x ^{2}}{a ^{2}} = 1$
Transverse axis	$2 a$ , along $x$ -axis	$2 b$ , along $y$ -axis
Conjugate axis	$2 b$ , along $y$ -axis	$2 a$ , along $x$ -axis
Relation	$b^{2} = a^{2} (e^{2} - 1)$	$a^{2} = b^{2} (e^{2} - 1)$
Eccentricity	$e = 1 + \frac{b ^{2}}{a ^{2}}$	$e = 1 + \frac{a ^{2}}{b ^{2}}$
Vertices	$(\pm a, 0)$	$(0, \pm b)$
Foci	$(\pm a e, 0)$	$(0, \pm b e)$
Directrices	$x = \pm \frac{a}{e}$	$y = \pm \frac{b}{e}$
Latus rectum	$\frac{2 b ^{2}}{a}$	$\frac{2 a ^{2}}{b}$

Note:

(a) If $e_{1}$ and $e_{2}$ are the eccentricities of a hyperbola and its conjugate, then $\frac{1}{e _{1}^{2}} + \frac{1}{e _{2}^{2}} = 1$ . This follows because $e_{1}^{2} = \frac{a ^{2} + b ^{2}}{a ^{2}}$ and $e_{2}^{2} = \frac{a ^{2} + b ^{2}}{b ^{2}}$ .
(b) The foci of a hyperbola and its conjugate are concyclic and form the vertices of a square. All four foci are at distance $a^{2} + b^{2}$ from the centre.
(c) Two hyperbolas are said to be similar if they have the same eccentricity.
(d) Two similar hyperbolas are said to be equal if they have the same latus rectum.
(e) If a hyperbola is equilateral, then its conjugate hyperbola is also equilateral.

Solved Example 3

Find the lengths of the transverse axis and conjugate axis, the eccentricity, the coordinates of the foci and vertices, the length of the latus rectum and the equations of the directrices of the hyperbola

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

Solution:

Dividing by 144, the equation becomes $\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = - 1$ , which is of the form $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = - 1$ (a conjugate-type hyperbola). So $a^{2} = 9, b^{2} = 16$ , giving $a = 3, b = 4$ .

Length of transverse axis: $2 b = 8$
Length of conjugate axis: $2 a = 6$
Eccentricity: $e = 1 + \frac{a ^{2}}{b ^{2}} = 1 + \frac{9}{16} = \frac{5}{4}$
Foci: $(0, \pm b e)$ , that is, $(0, \pm 5)$
Vertices: $(0, \pm b)$ , that is, $(0, \pm 4)$
Length of latus rectum: $\frac{2 a ^{2}}{b} = \frac{2 ( 3 ) ^{2}}{4} = \frac{9}{2}$
Directrices: $y = \pm \frac{b}{e} = \pm \frac{4}{5/4}$ , that is, $y = \pm \frac{16}{5}$

5. Auxiliary Circle

The circle drawn with centre $C$ and the transverse axis as diameter is called the auxiliary circle of the hyperbola. For $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ its equation is

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

In the figure below, $N$ is the foot of the ordinate of a point $P$ on the hyperbola and $N Q$ is a tangent to the auxiliary circle. The points $P$ and $Q$ are called the corresponding points of the hyperbola and the auxiliary circle. Since $∠ C QN = 9 0^{\circ}$ and $C Q = a$ , we get $C N = a sec θ$ where $θ = ∠ QC N$ .

Figure 5: Auxiliary circle

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

. The points

P (a sec θ, b tan θ)

on the hyperbola and

Q (a cos θ, a sin θ)

on the circle are corresponding points.

6. Parametric Representation

The equations $x = a sec θ$ and $y = b tan θ$ together represent the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ , where $θ$ is a parameter.

This works because $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = sec^{2} θ - tan^{2} θ = 1$ . Values $θ \in (- \frac{π}{2}, \frac{π}{2})$ give the right branch and $θ \in (\frac{π}{2}, \frac{3 π}{2})$ give the left branch.

Note: If $P (θ) \equiv (a sec θ, b tan θ)$ is on the hyperbola, then its corresponding point $Q (θ) \equiv (a cos θ, a sin θ)$ is on the auxiliary circle. The angle $θ$ is called the eccentric angle of $P$ .

7. Position of a Point with Respect to a Hyperbola

For the point $(x_{1}, y_{1})$ , the quantity

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

is positive, zero or negative according as the point lies inside, on or outside the hyperbola. Here "inside" means the region between a branch and its focus.

Solved Example 4

Find the position of the point

(5, - 4)

relative to the hyperbola

9 x^{2} - y^{2} = 1

Solution:

Here $S_{1} = 9 (5)^{2} - (- 4)^{2} - 1 = 225 - 16 - 1 = 208 > 0$ .

So the point $(5, - 4)$ lies inside the hyperbola $9 x^{2} - y^{2} = 1$ .

8. Practice Problems

Find the equation of the hyperbola whose foci are $(6, 4)$ and $(- 4, 4)$ and eccentricity is 2.
Answer: $12 x^{2} - 4 y^{2} - 24 x + 32 y - 127 = 0$
Obtain the equation of a hyperbola with the coordinate axes as principal axes, given that the distances of one of its vertices from the foci are 9 and 1 units.
Answer: $\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1$ or $\frac{y ^{2}}{16} - \frac{x ^{2}}{9} = 1$
The foci of a hyperbola coincide with the foci of the ellipse $\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1$ . Find the equation of the hyperbola if its eccentricity is 2.
Answer: $3 x^{2} - y^{2} - 12 = 0$

Common Mistakes to Avoid

Watch out

Using the ellipse relation $e^{2} = 1 - \frac{b ^{2}}{a ^{2}}$ . For a hyperbola it is $e^{2} = 1 + \frac{b ^{2}}{a ^{2}}$ , so $e$ is always greater than 1.
Assuming $a > b$ . In a hyperbola $a$ can be smaller than, equal to or larger than $b$ . The transverse axis is decided by which squared term is positive, not by which denominator is larger.
For $\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = - 1$ , taking the transverse axis as $2 a = 6$ . The right-hand side is $- 1$ , so the transverse axis is along the $y$ -axis with length $2 b = 8$ .
Writing $e_{1}^{2} + e_{2}^{2} = 1$ for a hyperbola and its conjugate. The correct relation is $\frac{1}{e _{1}^{2}} + \frac{1}{e _{2}^{2}} = 1$ .
Reading $S_{1} > 0$ as "outside". For a hyperbola, $S_{1} > 0$ means the point lies inside, in the region containing a focus. This is the opposite sign convention to the ellipse.
Using $(a cos θ, b sin θ)$ as the parametric point. That belongs to the ellipse; the hyperbola uses $(a sec θ, b tan θ)$ .
Forgetting to add the centre in a shifted hyperbola: the foci of $\frac{( x - h ) ^{2}}{a ^{2}} - \frac{( y - k ) ^{2}}{b ^{2}} = 1$ are $(h \pm a e, k)$ , not $(\pm a e, 0)$ .

Frequently Asked Questions

What is a hyperbola in coordinate geometry?

A hyperbola is the locus of a point whose distance from a fixed point (focus) bears a constant ratio $e > 1$ to its distance from a fixed line (directrix). Equivalently, the difference of its distances from the two foci is constant and equal to $2 a$ . Its standard equation is $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ .

How do you find the eccentricity of a hyperbola?

For $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ , use $e = 1 + \frac{b ^{2}}{a ^{2}}$ , where $a$ is the semi-transverse axis. For the conjugate form $\frac{y ^{2}}{b ^{2}} - \frac{x ^{2}}{a ^{2}} = 1$ , use $e = 1 + \frac{a ^{2}}{b ^{2}}$ . The value always exceeds 1, and it equals $2$ when $a = b$ .

What is the difference between the transverse axis and the conjugate axis?

The transverse axis is the segment joining the two vertices; both foci lie on its line and its length is $2 a$ for $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ . The conjugate axis is the perpendicular segment through the centre joining $(0, \pm b)$ , with length $2 b$ . The curve never meets its conjugate axis.

What is the length of the latus rectum of a hyperbola?

The latus rectum is the focal chord perpendicular to the transverse axis. For $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ its length is $\frac{2 b ^{2}}{a}$ , which also equals $2 a (e^{2} - 1)$ and the square of the conjugate axis divided by the transverse axis. Its end points are $(\pm a e, \pm \frac{b ^{2}}{a})$ .

What is a conjugate hyperbola?

The conjugate of $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ is $\frac{y ^{2}}{b ^{2}} - \frac{x ^{2}}{a ^{2}} = 1$ , whose transverse and conjugate axes are swapped. Both share the same centre and asymptotes, their four foci form a square, and their eccentricities satisfy $\frac{1}{e _{1}^{2}} + \frac{1}{e _{2}^{2}} = 1$ .

Why does a hyperbola use sec and tan in its parametric form?

Substituting $x = a sec θ$ and $y = b tan θ$ into $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}}$ gives $sec^{2} θ - tan^{2} θ = 1$ , so every value of $θ$ gives a point on the curve. Geometrically, $θ$ is the angle at the centre to the corresponding point $(a cos θ, a sin θ)$ on the auxiliary circle.

Is the hyperbola in the JEE Main syllabus?

Yes. The JEE Main syllabus includes equations of the parabola, ellipse and hyperbola in standard form, and questions on eccentricity, foci, latus rectum and the conjugate hyperbola are common. Every result on this page is directly usable in JEE Main problems.

Which basic hyperbola results matter most for JEE Advanced?

JEE Advanced lists foci, directrices, eccentricity and parametric equations of the hyperbola, followed by tangents, normals and locus problems. Be fluent with $b^{2} = a^{2} (e^{2} - 1)$ , the focal distance property $∣ S^{'} P - S P ∣ = 2 a$ , the parametric point $(a sec θ, b tan θ)$ and the sign of $S_{1}$ , since multi-step problems build on them.

Previous year questions on Introduction to Hyperbola

24 questions from past papers, each with a step-by-step solution.

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1. Definition of a Hyperbola

Hyperbola as a conic section

2. Standard Equation of a Hyperbola

Basic terms of the hyperbola

Length of the latus rectum

Focal distances of a point

3. Hyperbola with Centre at (h,k)

4. Conjugate Hyperbola

5. Auxiliary Circle

6. Parametric Representation

7. Position of a Point with Respect to a Hyperbola

8. Practice Problems

Common Mistakes to Avoid

Frequently Asked Questions

What is a hyperbola in coordinate geometry?

How do you find the eccentricity of a hyperbola?

What is the difference between the transverse axis and the conjugate axis?

What is the length of the latus rectum of a hyperbola?

What is a conjugate hyperbola?

Why does a hyperbola use sec and tan in its parametric form?

Is the hyperbola in the JEE Main syllabus?

Which basic hyperbola results matter most for JEE Advanced?

Previous year questions on Introduction to Hyperbola

Ready to master Hyperbola?

3. Hyperbola with Centre at $(h, k)$