Hyperbola - The Open Conic with Asymptotes

Master hyperbola equations, asymptotes, rectangular hyperbola, conjugate hyperbola, tangent and normal for JEE Main and Advanced

The Hook: From Sonic Booms to LORAN Navigation

Connect: Real Life → Mathematics

Ever seen a sonic boom? When a jet breaks the sound barrier, it creates a hyperbolic shock wave cone!

Real-World Hyperbolas:

  • Sonic booms - Supersonic aircraft create hyperbolic shock waves
  • LORAN navigation - Ships and aircraft use hyperbolic positioning
  • Cooling towers of nuclear power plants - hyperbolic shape for strength
  • Shadows of cones on walls - hyperbolic curves
  • Radio telescopes - hyperbolic reflectors
  • Comets with escape velocity - hyperbolic orbits around the Sun

From Physics: When a particle approaches light speed, its energy-momentum relation forms a hyperbola!

The Question: What makes a hyperbola different from an ellipse? Why do some comets follow hyperbolic paths while planets follow elliptical ones?

Answer: The focal difference property - the difference of distances from any point to two foci is constant (ellipse uses sum, hyperbola uses difference). This creates an open curve that never closes!

JEE Impact: Hyperbola appears in 2-3 questions every year in JEE Main and 1-2 in Advanced. Combined with rectangular hyperbola, it’s a high-scoring topic!


The Core Concept

A hyperbola is the locus of all points such that the absolute difference of distances from two fixed points (foci) is constant.

The Big Idea

Understanding Hyperbola

Mathematical Definition:

$$|PF_1 - PF_2| = 2a \quad \text{(constant)}$$

where $P$ is any point on the hyperbola, and $F_1$, $F_2$ are the foci.

Comparison with Ellipse:

  • Ellipse: SUM of focal distances = $2a$ → Closed curve
  • Hyperbola: DIFFERENCE of focal distances = $2a$ → Open curve

Visual Description:

  • Two separate branches (unlike ellipse which is one closed curve)
  • Opens infinitely in opposite directions
  • Has asymptotes - straight lines that the curve approaches but never touches
  • More “spread out” than ellipse
  • Symmetric about both transverse and conjugate axes

Think of it as: Two mirror-image parabolas facing away from each other, but with different mathematical properties.


Standard Forms of Hyperbola

1. Horizontal Hyperbola (Transverse Axis Along X-axis)

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

Interactive Demo: Visualize Hyperbolas

Standard Hyperbola: x²/a² - y²/b² = 1XYabcCenter O(0,0)A(a, 0)A'(-a, 0)F1(-c, 0)F2(c, 0)y = (b/a)xy = -(b/a)xPTransverse = 2aConjugate = 2bKey Relations:c² = a² + b² (ADD!)e = c/a (e > 1)Focal Property:|PF1 - PF2| = 2a(Difference is constant)Legend:FociVerticesAsymptotes

Explore all conic sections interactively! For hyperbola, drag the point P to verify that |PF1 - PF2| = 2a (the difference of distances to both foci is constant). Toggle asymptotes to see how the curve approaches but never touches these lines.

Key Elements:

  • Center: $(0, 0)$
  • Vertices: $(\pm a, 0)$ on x-axis
  • Foci: $(\pm c, 0)$ where $c = \sqrt{a^2 + b^2}$
  • Transverse axis: Along x-axis, length $= 2a$
  • Conjugate axis: Along y-axis, length $= 2b$
  • Eccentricity: $e = \frac{c}{a} = \frac{\sqrt{a^2 + b^2}}{a}$ where $e > 1$
  • Asymptotes: $y = \pm \frac{b}{a}x$

Visual Description: Opens left and right, like two parabolas back-to-back horizontally.

2. Vertical Hyperbola (Transverse Axis Along Y-axis)

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

Key Elements:

  • Center: $(0, 0)$
  • Vertices: $(0, \pm a)$ on y-axis
  • Foci: $(0, \pm c)$ where $c = \sqrt{a^2 + b^2}$
  • Transverse axis: Along y-axis, length $= 2a$
  • Conjugate axis: Along x-axis, length $= 2b$
  • Eccentricity: $e = \frac{c}{a} > 1$
  • Asymptotes: $y = \pm \frac{a}{b}x$

Visual Description: Opens up and down vertically.

How to Identify the Opening Direction

'Positive Term Opens There'

Simple Rule:

The POSITIVE term tells you the opening direction!

  • If $\frac{x^2}{a^2}$ is positive → Opens horizontally (left-right)
  • If $\frac{y^2}{a^2}$ is positive → Opens vertically (up-down)

Key Difference from Ellipse:

  • Ellipse: BOTH terms positive ($+$ and $+$)
  • Hyperbola: ONE term positive, ONE negative ($+$ and $-$)

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

  • $x^2$ term positive → Horizontal hyperbola
  • $a^2 = 16$ → $a = 4$, $b^2 = 9$ → $b = 3$

Example 2: $\frac{y^2}{25} - \frac{x^2}{16} = 1$

  • $y^2$ term positive → Vertical hyperbola
  • $a^2 = 25$ → $a = 5$, $b^2 = 16$ → $b = 4$

Important Terms and Definitions

1. Transverse Axis

The axis passing through the foci and vertices.

Length: $2a$

This is where the hyperbola actually exists (the vertices are on this axis).

2. Conjugate Axis

The axis perpendicular to the transverse axis.

Length: $2b$

Note: Unlike ellipse, the hyperbola does not intersect the conjugate axis!

3. Foci

Two fixed points, one on each branch.

Location: At distance $c$ from center along transverse axis

Key Relation: $c^2 = a^2 + b^2$ (ADDITION!)

Remember: $c > a$ (foci are farther from center than vertices)

4. Eccentricity ($e$)

A measure of how “open” the hyperbola is.

$$\boxed{e = \frac{c}{a} = \frac{\sqrt{a^2 + b^2}}{a}}$$

Range: $e > 1$ (always greater than 1)

Special Cases:

  • $e$ close to 1 → Narrow hyperbola (looks like two parabolas)
  • $e$ large → Very open hyperbola (branches more spread out)
  • $e = \sqrt{2}$ → Rectangular hyperbola

Alternative Forms:

$$\boxed{b^2 = a^2(e^2 - 1)}$$ $$\boxed{c^2 = a^2 e^2}$$

5. Latus Rectum

A chord through the focus, perpendicular to the transverse axis.

Length: $\frac{2b^2}{a}$ (same formula as ellipse!)

Endpoints for horizontal hyperbola: $\left(c, \pm\frac{b^2}{a}\right)$

6. Asymptotes

Straight lines that the hyperbola approaches but never touches as it extends to infinity.

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

$$\boxed{y = \pm \frac{b}{a}x}$$

Or equivalently: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 0$

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

$$\boxed{y = \pm \frac{a}{b}x}$$

Visual Description: Imagine two diagonal lines crossing at the center. As you move farther from the center along the hyperbola, the curve gets closer and closer to these lines but never actually reaches them.

Why Asymptotes Matter

Asymptotes are the skeleton of the hyperbola!

  1. They pass through the center
  2. They divide the plane into four regions
  3. The hyperbola exists only in two opposite regions
  4. Drawing asymptotes first makes sketching hyperbola easy

Quick Sketch Method:

  1. Draw asymptotes $y = \pm \frac{b}{a}x$
  2. Mark vertices at $(\pm a, 0)$
  3. Draw smooth curves from vertices approaching asymptotes

7. Directrix

Lines associated with each focus.

For horizontal hyperbola:

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

Property: For any point $P$ on hyperbola:

$$\frac{\text{Distance from P to focus}}{\text{Distance from P to directrix}} = e$$

Key Relations (Memory Gems!)

$$\boxed{c^2 = a^2 + b^2} \quad \text{(ADDITION - opposite of ellipse!)}$$ $$\boxed{e = \frac{c}{a} > 1}$$ $$\boxed{b^2 = a^2(e^2 - 1)}$$ $$\boxed{\text{Asymptotes: } y = \pm \frac{b}{a}x \text{ (for horizontal)}}$$ $$\boxed{\text{Latus Rectum} = \frac{2b^2}{a}}$$
Memory Trick: 'Hyperbola ADDS'

“Ellipse SUBTRACTS, Hyperbola ADDS”

Ellipse: $c^2 = a^2 - b^2$ (subtraction)

Hyperbola: $c^2 = a^2 + b^2$ (addition)

Why?

  • Ellipse: Foci are inside (closer than semi-major axis) → subtract
  • Hyperbola: Foci are outside (farther than semi-transverse axis) → add

Visual: Think of the right triangle:

  • Vertices at $(\pm a, 0)$
  • Asymptote passes through $(a, b)$ from origin
  • Focus at $(c, 0)$
  • Triangle: $(0,0)$, $(a, 0)$, $(a, b)$ → hypotenuse length = $\sqrt{a^2 + b^2} = c$

Rectangular Hyperbola

A rectangular hyperbola is a special hyperbola where:

$$\boxed{a = b}$$

or equivalently:

$$\boxed{e = \sqrt{2}}$$

Standard Form

$$\boxed{x^2 - y^2 = a^2}$$

Asymptotes: $y = \pm x$ (perpendicular to each other at 45°)

Rotated Form (Most Common in JEE!)

When rotated 45°, the rectangular hyperbola becomes:

$$\boxed{xy = c^2}$$

or more generally:

$$\boxed{xy = k}$$

This is the equation you’ll see most often in JEE!

Key Properties:

  • Asymptotes: $x = 0$ and $y = 0$ (the coordinate axes!)
  • Eccentricity: $e = \sqrt{2}$
  • Parametric form: $x = ct$, $y = \frac{c}{t}$ (where $t \neq 0$)

Visual Description: The familiar “reciprocal curve” you see in $y = \frac{1}{x}$ graphs. Two branches in opposite quadrants, approaching both axes but never touching them.

Example: Rectangular Hyperbola in Real Life

Boyle’s Law in Chemistry: $PV = k$ (constant temperature)

  • Pressure $P$ vs Volume $V$ graph is a rectangular hyperbola!
  • As volume increases, pressure decreases
  • Product remains constant

Economics: Demand-Price relationship often follows $xy = k$ form.


Conjugate Hyperbola

For every hyperbola, there exists a conjugate hyperbola.

If hyperbola is: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

Conjugate hyperbola is: $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$

Or simply: $-\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

Key Properties:

  1. Same asymptotes as original hyperbola
  2. Transverse and conjugate axes are interchanged
  3. Original hyperbola opens horizontally, conjugate opens vertically (or vice versa)
  4. They are reflections about the asymptotes

Visual Description: If you draw both on the same graph, they share the same asymptotes but open in perpendicular directions, filling all four regions created by the asymptotes.


Parametric Equations

For hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$:

$$\boxed{x = a\sec\theta, \quad y = b\tan\theta}$$

Range: $\theta \in [0, 2\pi)$, $\theta \neq \frac{\pi}{2}, \frac{3\pi}{2}$

Verification:

$$\frac{a^2\sec^2\theta}{a^2} - \frac{b^2\tan^2\theta}{b^2} = \sec^2\theta - \tan^2\theta = 1$$

For rectangular hyperbola $xy = c^2$:

$$\boxed{x = ct, \quad y = \frac{c}{t}} \quad \text{where } t \neq 0$$
Example: Parametric Point

Find point on hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$ at $\theta = 60°$.

Solution:

$a = 4$, $b = 3$

$$x = 4\sec 60° = 4 \times 2 = 8$$ $$y = 3\tan 60° = 3\sqrt{3}$$

Point: $(8, 3\sqrt{3})$

Verification: $\frac{64}{16} - \frac{27}{9} = 4 - 3 = 1$ ✓


Tangent to a Hyperbola

1. Tangent at Point $(x_1, y_1)$

For hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$:

$$\boxed{\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1}$$

2. Tangent at Parametric Point $(a\sec\theta, b\tan\theta)$

$$\boxed{\frac{x\sec\theta}{a} - \frac{y\tan\theta}{b} = 1}$$

3. Tangent with Slope $m$

$$\boxed{y = mx \pm \sqrt{a^2m^2 - b^2}}$$

Condition: For real tangent, $a^2m^2 - b^2 \geq 0$ → $m^2 \geq \frac{b^2}{a^2}$

This means: Not all slopes give tangents! Only slopes with $|m| \geq \frac{b}{a}$ work.

For rectangular hyperbola $xy = c^2$ at point $(ct, \frac{c}{t})$:

$$\boxed{\frac{x}{t} + ty = 2c}$$
Memory Trick: Same Pattern as Ellipse

For tangent at $(x_1, y_1)$:

Just like ellipse, replace $x^2$ with $xx_1$, $y^2$ with $yy_1$:

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

Tangent: $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$

Only difference: Keep the minus sign!

Example: Tangent Equation

Find equation of tangent to $\frac{x^2}{9} - \frac{y^2}{4} = 1$ at point $(5, \frac{8}{3})$.

Solution:

Using $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$:

$$\frac{x \cdot 5}{9} - \frac{y \cdot \frac{8}{3}}{4} = 1$$ $$\frac{5x}{9} - \frac{8y}{12} = 1$$ $$\frac{5x}{9} - \frac{2y}{3} = 1$$

Multiply by 9:

$$\boxed{5x - 6y = 9}$$

Normal to a Hyperbola

1. Normal at Point $(x_1, y_1)$

For hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$:

$$\boxed{\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2 = c^2}$$

2. Normal at Parametric Point $(a\sec\theta, b\tan\theta)$

$$\boxed{ax\cos\theta + by\cot\theta = a^2 + b^2}$$

For rectangular hyperbola $xy = c^2$ at $(ct, \frac{c}{t})$:

$$\boxed{xt^3 - yt - ct^4 + c = 0}$$

Important Properties of Hyperbola

1. Focal Difference Property

For any point $P$ on the hyperbola:

$$\boxed{|PF_1 - PF_2| = 2a}$$

This is the defining property!

Visual Description: No matter where you are on the hyperbola, the absolute difference of distances to both foci equals the transverse axis length.

2. Asymptotes are NOT Tangents

Key Point: Asymptotes touch the hyperbola “at infinity” but are NOT tangents in the usual sense.

Tangent touches at one point (finite), asymptote approaches but never actually touches.

3. Chord of Contact

From external point $(x_1, y_1)$, the chord joining two tangent points:

$$\boxed{\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1}$$

(Same form as tangent equation!)

4. Auxiliary Circle

Circle with radius $a$ (semi-transverse axis).

Equation: $x^2 + y^2 = a^2$

Connection: Helps in deriving parametric equations using $\sec\theta$ and $\tan\theta$.

5. Reflection Property

Property: A ray directed towards one focus, after reflecting off the hyperbola, appears to come from the other focus.

Application: LORAN navigation system uses this property - by measuring time difference of signals from two stations, position lies on a hyperbola.


Memory Tricks & Patterns

Pattern 1: Hyperbola vs Ellipse

PropertyEllipseHyperbola
Equation sign$+$ and $+$$+$ and $-$
Focal relation$c^2 = a^2 - b^2$$c^2 = a^2 + b^2$
Eccentricity$0 < e < 1$$e > 1$
Focal propertySum = $2a$Difference = $2a$
ShapeClosed curveOpen curve
AsymptotesNoneTwo asymptotes

Pattern 2: Finding Asymptotes

“Change 1 to 0”

From hyperbola equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$:

Replace $1$ with $0$: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 0$

This factors: $\left(\frac{x}{a} - \frac{y}{b}\right)\left(\frac{x}{a} + \frac{y}{b}\right) = 0$

Asymptotes: $y = \frac{b}{a}x$ and $y = -\frac{b}{a}x$

Pattern 3: Rectangular Hyperbola Recognition

Any equation of the form $xy = k$ is a rectangular hyperbola!

  • $xy = 1$ → Rectangular hyperbola
  • $xy = 5$ → Rectangular hyperbola
  • $xy = -3$ → Rectangular hyperbola (in different quadrants)

Common Mistakes to Avoid

Trap #1: Focus Formula Sign

Wrong: $c^2 = a^2 - b^2$ (ellipse formula)

Correct: $c^2 = a^2 + b^2$ (hyperbola formula)

Remember: “Hyperbola ADDS, Ellipse SUBTRACTS”

Why? Foci in hyperbola are farther from center than vertices ($c > a$), so we add.

Trap #2: Eccentricity Range

Wrong: Eccentricity of hyperbola is less than 1

Correct: For hyperbola, $e > 1$ ALWAYS

Remember:

  • Circle: $e = 0$
  • Ellipse: $0 < e < 1$
  • Parabola: $e = 1$
  • Hyperbola: $e > 1$
Trap #3: Asymptotes are NOT Tangents

Common Error: Treating asymptotes as tangent lines

Correct:

  • Tangent touches hyperbola at one finite point
  • Asymptote approaches hyperbola but never touches (meets at infinity)

Key: Asymptotes are limiting positions, not actual tangents.

Trap #4: Rectangular Hyperbola Condition

Wrong: Rectangular hyperbola means $a = b = 1$

Correct: Rectangular hyperbola means $a = b$ (can be any equal value) OR $e = \sqrt{2}$

Examples:

  • $x^2 - y^2 = 4$ → Rectangular hyperbola ($a = b = 2$)
  • $xy = 5$ → Rectangular hyperbola (standard rectangular form)

Practice Problems

Level 1: Foundation (NCERT Style)

Problem 1: Basic Elements

Question: For hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$, find vertices, foci, eccentricity, and asymptotes.

Solution:

$a^2 = 16$ → $a = 4$, $b^2 = 9$ → $b = 3$

Horizontal hyperbola (positive $x^2$ term)

$c^2 = a^2 + b^2 = 16 + 9 = 25$ → $c = 5$

  • Vertices: $(\pm 4, 0)$
  • Foci: $(\pm 5, 0)$
  • Eccentricity: $e = \frac{c}{a} = \frac{5}{4}$
  • Asymptotes: $y = \pm \frac{3}{4}x$
  • Latus Rectum: $\frac{2b^2}{a} = \frac{2 \times 9}{4} = \frac{9}{2}$
Problem 2: Finding Equation

Question: Find equation of hyperbola with foci at $(\pm 5, 0)$ and vertices at $(\pm 3, 0)$.

Solution:

Foci and vertices on x-axis → horizontal hyperbola

$c = 5$, $a = 3$

$b^2 = c^2 - a^2 = 25 - 9 = 16$ → $b = 4$

$$\boxed{\frac{x^2}{9} - \frac{y^2}{16} = 1}$$
Problem 3: Rectangular Hyperbola Point

Question: Find point on rectangular hyperbola $xy = 8$ with parameter $t = 2$.

Solution:

For $xy = c^2$, we have $c^2 = 8$ → $c = 2\sqrt{2}$

Parametric: $x = ct = 2\sqrt{2} \times 2 = 4\sqrt{2}$

$y = \frac{c}{t} = \frac{2\sqrt{2}}{2} = \sqrt{2}$

Point: $(4\sqrt{2}, \sqrt{2})$

Verification: $4\sqrt{2} \times \sqrt{2} = 4 \times 2 = 8$ ✓

Level 2: JEE Main Type

Problem 4: Focal Difference

Question: A point on hyperbola $\frac{x^2}{25} - \frac{y^2}{16} = 1$ is at distance 7 from one focus. Find its distance from the other focus.

Solution:

$a^2 = 25$ → $a = 5$

By focal difference property: $|PF_1 - PF_2| = 2a = 10$

Case 1: $PF_1 - PF_2 = 10$

If $PF_1 = 7$: $7 - PF_2 = 10$ → $PF_2 = -3$ (impossible)

Case 2: $PF_2 - PF_1 = 10$

If $PF_1 = 7$: $PF_2 - 7 = 10$ → $PF_2 = 17$

$$\boxed{PF_2 = 17}$$
Problem 5: Tangent with Given Slope

Question: Find equations of tangents to $\frac{x^2}{16} - \frac{y^2}{9} = 1$ with slope $2$.

Solution:

$a^2 = 16$, $b^2 = 9$, $m = 2$

Check condition: $a^2m^2 - b^2 = 16(4) - 9 = 64 - 9 = 55 > 0$ ✓

Using $y = mx \pm \sqrt{a^2m^2 - b^2}$:

$$y = 2x \pm \sqrt{55}$$ $$\boxed{y = 2x + \sqrt{55} \quad \text{and} \quad y = 2x - \sqrt{55}}$$

Or: $2x - y \pm \sqrt{55} = 0$

Problem 6: Conjugate Hyperbola

Question: Find equation of conjugate hyperbola of $\frac{x^2}{9} - \frac{y^2}{16} = 1$.

Solution:

Conjugate: Interchange $x^2$ and $y^2$ terms and change sign:

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

Or equivalently: $-\frac{x^2}{9} + \frac{y^2}{16} = 1$

Note: Both have same asymptotes $y = \pm \frac{4}{3}x$

Level 3: JEE Advanced Type

Problem 7: Locus Problem

Question: Find locus of midpoint of chords of hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ parallel to line $y = mx$.

Solution:

Let midpoint be $(h, k)$.

For chord parallel to $y = mx$, the chord has slope $m$.

Using T = S₁ method:

Tangent at $(h, k)$: $\frac{xh}{a^2} - \frac{yk}{b^2} = \frac{h^2}{a^2} - \frac{k^2}{b^2}$

Slope of this tangent: $\frac{b^2h}{a^2k}$

For chord through $(h, k)$ with slope $m$:

Slope = $m$ = $\frac{b^2h}{a^2k}$

$$b^2h = ma^2k$$

Locus (replace $(h, k)$ with $(x, y)$):

$$\boxed{b^2x = ma^2y}$$

Or: $b^2x - ma^2y = 0$

This is a straight line through the origin!

Problem 8: Director Circle

Question: Find locus of point of intersection of perpendicular tangents to $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

Solution:

This is the director circle.

For hyperbola:

$$\boxed{x^2 + y^2 = a^2 - b^2}$$

Note: This exists only when $a^2 > b^2$ (real circle).

Problem 9: Rectangular Hyperbola Tangent

Question: Find equation of tangent to $xy = 16$ at point with parameter $t = 4$.

Solution:

$xy = c^2 = 16$ → $c = 4$

Point: $(ct, \frac{c}{t}) = (4 \times 4, \frac{4}{4}) = (16, 1)$

Using tangent formula for rectangular hyperbola:

$$\frac{x}{t} + ty = 2c$$ $$\frac{x}{4} + 4y = 2(4)$$ $$\frac{x}{4} + 4y = 8$$

Multiply by 4:

$$\boxed{x + 16y = 32}$$

Verification: Point $(16, 1)$ satisfies: $16 + 16 = 32$ ✓


Quick Revision Box

SituationFormula/Approach
Horizontal hyperbola$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
Vertical hyperbola$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$
Focus relation$c^2 = a^2 + b^2$ (ADDITION)
Eccentricity$e = \frac{c}{a} > 1$
Focal difference$
Asymptotes$y = \pm \frac{b}{a}x$ (horizontal)
Rectangular hyperbola$xy = c^2$ or $a = b$
Rectangular $e$$e = \sqrt{2}$
Parametric (standard)$x = a\sec\theta$, $y = b\tan\theta$
Parametric (rect.)$x = ct$, $y = \frac{c}{t}$
Tangent at $(x_1,y_1)$$\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$
Latus rectum$\frac{2b^2}{a}$
Conjugate hyperbolaInterchange and negate

Prerequisites:

Related Topics:

Applications:

  • Physics: Particle trajectories, sonic boom patterns
  • Navigation: LORAN positioning system
  • Engineering: Cooling tower design, reflector shapes
  • Astronomy: Hyperbolic comet orbits

JEE Exam Strategy

High-Yield Points for JEE

Weightage: 2-3 questions in JEE Main, 1-2 in Advanced

Most Frequently Asked:

  1. Rectangular hyperbola $xy = c^2$ - very popular in JEE!
  2. Finding equation given conditions (foci, vertices, eccentricity, asymptotes)
  3. Focal difference property $|PF_1 - PF_2| = 2a$
  4. Asymptote equations and their properties
  5. Tangent and normal equations

Time-Saving Tricks:

  1. For rectangular hyperbola: Use $xy = c^2$ and parametric $x = ct, y = \frac{c}{t}$ directly
  2. Asymptotes: Change “= 1” to “= 0” in hyperbola equation
  3. Remember: $c^2 = a^2 + b^2$ (ADDITION, not subtraction!)
  4. Eccentricity check: Must be $> 1$ for hyperbola

Common Trap Options:

  • Giving $c^2 = a^2 - b^2$ (ellipse formula) instead of $c^2 = a^2 + b^2$
  • Wrong eccentricity range ($< 1$ instead of $> 1$)
  • Treating asymptotes as tangents
  • Forgetting absolute value in focal difference: $|PF_1 - PF_2| = 2a$

Pattern Recognition:

  • "$xy = k$" → Rectangular hyperbola, use parametric immediately
  • “Asymptotes” → Change equation to “= 0” form
  • “Perpendicular asymptotes” → Rectangular hyperbola ($a = b$)
  • “Conjugate hyperbola” → Swap positive and negative terms

Quick Checks:

  • Verify $e > 1$ always for hyperbola
  • Check $c > a$ (foci farther than vertices)
  • Asymptotes pass through center
  • For rectangular hyperbola: verify $a = b$ or $e = \sqrt{2}$

Teacher’s Summary

Key Takeaways
  1. Definition: Difference of distances from any point to two foci equals $2a$ (constant)
  2. Key relation: $c^2 = a^2 + b^2$ (ADDITION - opposite of ellipse!)
  3. Open curve: Two separate branches approaching asymptotes
  4. Asymptotes: Lines that hyperbola approaches at infinity: $y = \pm \frac{b}{a}x$
  5. Eccentricity: $e > 1$ always (measures how “open” the hyperbola is)
  6. Rectangular hyperbola: $a = b$ (equivalently $e = \sqrt{2}$), often written as $xy = c^2$
  7. Conjugate hyperbola: Opens in perpendicular direction, shares same asymptotes
  8. Reflection property: Ray towards one focus reflects as if from other focus (LORAN navigation)

“The hyperbola is the open conic - while ellipse closes around its foci, hyperbola spreads infinitely, forever approaching but never touching its asymptotes!”

Mastery Check: Can you instantly:

  • Write asymptote equations from hyperbola equation?
  • Identify if $xy = k$ is a hyperbola? (Yes - rectangular!)
  • Remember $c^2 = a^2 + b^2$ (not minus)?
  • Recall $e > 1$ for hyperbola?

Connection to Other Conics:

  • Circle: $e = 0$ (both foci coincide)
  • Ellipse: $0 < e < 1$ (sum of focal distances constant)
  • Parabola: $e = 1$ (one focus, one directrix)
  • Hyperbola: $e > 1$ (difference of focal distances constant)

You’ve completed all major conics! You now understand the complete family of curves obtained by slicing a cone. Master these, and coordinate geometry is yours!


What’s Next?

You’ve now mastered all five topics in Coordinate Geometry:

  1. Straight Lines - The foundation
  2. Circles - The perfect curve
  3. Parabola - The focusing curve
  4. Ellipse - The orbital curve
  5. Hyperbola - The open curve

Continue your JEE preparation with:

Practice Integration: Many JEE problems combine multiple topics:

  • Circle and straight line intersection
  • Ellipse and tangent properties
  • Parabola focus and reflection
  • All conics together in one problem

Keep practicing, and remember: Coordinate Geometry is one of the highest-scoring topics in JEE if you master the formulas and patterns!