The Hook: From Satellite Dishes to Football Kicks
Look up at any satellite dish on a rooftop! Why is it shaped like a parabola?
- Satellite dishes focus all incoming signals to one point
- Car headlights use parabolic reflectors to create parallel beams
- Football kicks and water fountains follow parabolic paths
- Solar cookers use parabolic mirrors to concentrate sunlight
From “Pathaan”: When Shah Rukh Khan jumps from a plane or a vehicle takes a projectile path - that’s a parabola!
The Question: What makes a parabola so special that it appears everywhere from space technology to sports?
Answer: The focus-directrix property - every point on a parabola is equidistant from a fixed point (focus) and a fixed line (directrix). This creates the perfect reflecting curve!
JEE Impact: Parabola appears in 3-4 questions every year in JEE Main and 2-3 in Advanced. It’s a high-scoring topic with predictable patterns!
The Core Concept
A parabola is the locus of all points that are equidistant from a fixed point (focus) and a fixed line (directrix).
The Big Idea
Imagine this:
- You have a point F (focus) and a line (directrix)
- You want to find all points P such that:
- Distance from P to F = Distance from P to the directrix line
This collection of all such points P forms a parabola!
Visual Description:
- The parabola “wraps around” the focus
- It opens away from the directrix
- The vertex is exactly halfway between focus and directrix
- The axis of symmetry passes through focus and is perpendicular to directrix
Interactive Demo: Parabola Visualization
Explore all conic sections interactively! Switch between Circle, Ellipse, Parabola, and Hyperbola to see how they relate. Drag the point P to see the focus-directrix property in action, where PF/PM = e (eccentricity).
Try plotting parabolas with different equations: $y^2 = 4ax$, $x^2 = 4ay$, $(y-k)^2 = 4a(x-h)$. See how the focus, directrix, and vertex positions change!
Standard Forms of Parabola
1. Parabola Opening Rightward (Most Common)
$$\boxed{y^2 = 4ax}$$Where:
- Vertex: $(0, 0)$
- Focus: $(a, 0)$ on positive x-axis
- Directrix: $x = -a$ (vertical line)
- Axis: $y = 0$ (x-axis)
- Latus Rectum: $4a$
Visual Description: Opens towards the right, like a satellite dish mounted on a vertical wall.
2. Parabola Opening Leftward
$$\boxed{y^2 = -4ax}$$Where:
- Vertex: $(0, 0)$
- Focus: $(-a, 0)$ on negative x-axis
- Directrix: $x = a$
- Axis: $y = 0$
- Latus Rectum: $4a$
Visual Description: Opens towards the left, mirror image of the rightward parabola.
3. Parabola Opening Upward
$$\boxed{x^2 = 4ay}$$Where:
- Vertex: $(0, 0)$
- Focus: $(0, a)$ on positive y-axis
- Directrix: $y = -a$
- Axis: $x = 0$ (y-axis)
- Latus Rectum: $4a$
Visual Description: Opens upward like a water fountain or football kick trajectory.
4. Parabola Opening Downward
$$\boxed{x^2 = -4ay}$$Where:
- Vertex: $(0, 0)$
- Focus: $(0, -a)$ on negative y-axis
- Directrix: $y = a$
- Axis: $x = 0$
- Latus Rectum: $4a$
Visual Description: Opens downward like an inverted dome or bridge arch.
Summary Table
| Equation | Vertex | Focus | Directrix | Axis | Opens |
|---|---|---|---|---|---|
| $y^2 = 4ax$ | $(0,0)$ | $(a,0)$ | $x = -a$ | $y=0$ | Right |
| $y^2 = -4ax$ | $(0,0)$ | $(-a,0)$ | $x = a$ | $y=0$ | Left |
| $x^2 = 4ay$ | $(0,0)$ | $(0,a)$ | $y = -a$ | $x=0$ | Up |
| $x^2 = -4ay$ | $(0,0)$ | $(0,-a)$ | $y = a$ | $x=0$ | Down |
“For parabola, focus on the ‘a’”
Rule: The variable that is squared tells you the direction:
- $y^2$ = parabola opens horizontally (left or right)
- $x^2$ = parabola opens vertically (up or down)
The sign tells direction:
- Positive $4a$ = opens in positive direction (right or up)
- Negative $-4a$ = opens in negative direction (left or down)
Example: $y^2 = 4ax$
- $y$ is squared → horizontal parabola
- Positive → opens right
- Focus at $(a, 0)$ → “a” units on the positive x-axis
Important Terms and Definitions
1. Vertex
The point on the parabola closest to the directrix. For standard parabolas, it’s at the origin $(0, 0)$.
2. Focus
A fixed point from which distances are measured. All parabolic reflectors reflect to/from this point.
Distance from vertex to focus: Always $a$
3. Directrix
A fixed line. Every point on the parabola is equidistant from the focus and this line.
Distance from vertex to directrix: Always $a$
4. Axis of Symmetry
The line passing through the focus and perpendicular to the directrix. The parabola is symmetric about this line.
5. Latus Rectum
The chord passing through the focus and perpendicular to the axis.
Length: Always $4a$
Endpoints for $y^2 = 4ax$: $(a, 2a)$ and $(a, -2a)$
Why Important? The latus rectum determines the “width” of the parabola at the focus.
For parabola $y^2 = 12x$, find vertex, focus, directrix, and latus rectum.
Solution:
Compare with $y^2 = 4ax$:
$$4a = 12 \implies a = 3$$- Vertex: $(0, 0)$
- Focus: $(3, 0)$
- Directrix: $x = -3$
- Axis: $y = 0$
- Latus Rectum: $4a = 12$
- Endpoints of LR: $(3, 6)$ and $(3, -6)$
General Parabola (Vertex Not at Origin)
Vertex at $(h, k)$
Horizontal axis (opens right):
$$\boxed{(y - k)^2 = 4a(x - h)}$$- Vertex: $(h, k)$
- Focus: $(h + a, k)$
- Directrix: $x = h - a$
Vertical axis (opens up):
$$\boxed{(x - h)^2 = 4a(y - k)}$$- Vertex: $(h, k)$
- Focus: $(h, k + a)$
- Directrix: $y = k - a$
Find vertex and focus of $(y - 2)^2 = 8(x - 1)$.
Solution:
Compare with $(y - k)^2 = 4a(x - h)$:
$k = 2$, $h = 1$, $4a = 8$ → $a = 2$
- Vertex: $(1, 2)$
- Focus: $(1 + 2, 2) = (3, 2)$
- Directrix: $x = 1 - 2 = -1$
Parametric Equations
For parabola $y^2 = 4ax$, any point can be written as:
$$\boxed{x = at^2, \quad y = 2at}$$where $t$ is the parameter.
Why Parametric?
- Easier to find coordinates of points on parabola
- Simplifies tangent and normal equations
- Useful in calculus and integration
Verification:
$$y^2 = (2at)^2 = 4a^2t^2 = 4a(at^2) = 4ax$$✓
Find the point on parabola $y^2 = 16x$ with parameter $t = 2$.
Solution:
Here $4a = 16$ → $a = 4$
$$x = 4(2)^2 = 16$$ $$y = 2(4)(2) = 16$$Point: $(16, 16)$
Verification: $16^2 = 256 = 16(16)$ ✓
Tangent to a Parabola
1. Tangent at Point $(x_1, y_1)$
For $y^2 = 4ax$:
$$\boxed{yy_1 = 2a(x + x_1)}$$2. Tangent at Parametric Point $(at^2, 2at)$
$$\boxed{ty = x + at^2}$$This is the most useful form for JEE!
3. Tangent with Slope $m$
$$\boxed{y = mx + \frac{a}{m}}$$Note: For a given slope $m$, there is exactly one tangent to the parabola (unlike circles which have two).
Point of contact: $\left(\frac{a}{m^2}, \frac{2a}{m}\right)$
For tangent at $(x_1, y_1)$ on $y^2 = 4ax$:
Replace: $y^2$ with $yy_1$ and $x$ with $\frac{x + x_1}{2}$
From $y^2 = 4ax$:
Tangent: $yy_1 = 4a \cdot \frac{x + x_1}{2}$
$$yy_1 = 2a(x + x_1)$$Parametric form is even simpler: Just $ty = x + at^2$
Find equation of tangent to $y^2 = 8x$ at point $(2, 4)$.
Solution:
Here $4a = 8$ → $a = 2$
Using $yy_1 = 2a(x + x_1)$:
$$y(4) = 2(2)(x + 2)$$ $$4y = 4x + 8$$ $$\boxed{y = x + 2}$$Alternate (parametric): At $(2, 4)$, find $t$:
$x = 2t^2 = 2$ → $t^2 = 1$ → $t = 2$ (taking positive)
Wait, let me recalculate: $y = 2at = 2(2)t = 4t = 4$ → $t = 1$
Using $ty = x + at^2$:
$$1 \cdot y = x + 2(1)^2$$ $$\boxed{y = x + 2}$$✓
Normal to a Parabola
1. Normal at Point $(x_1, y_1)$
For $y^2 = 4ax$:
$$\boxed{y - y_1 = -\frac{y_1}{2a}(x - x_1)}$$2. Normal at Parametric Point $(at^2, 2at)$
$$\boxed{y + tx = 2at + at^3}$$This is the most useful form!
3. Normal with Slope $m$
$$\boxed{y = mx - 2am - am^3}$$Note: For a given slope $m$, there can be up to 3 normals to the parabola (unlike tangent which is unique for given slope).
Find equation of normal to $y^2 = 12x$ at point $(3, 6)$.
Solution:
Here $4a = 12$ → $a = 3$
Using $y - y_1 = -\frac{y_1}{2a}(x - x_1)$:
$$y - 6 = -\frac{6}{2(3)}(x - 3)$$ $$y - 6 = -1(x - 3)$$ $$y - 6 = -x + 3$$ $$\boxed{x + y = 9}$$Properties of Focal Chord
A focal chord is a chord passing through the focus.
Key Properties
For parabola $y^2 = 4ax$:
If endpoints have parameters $t_1$ and $t_2$:
$$\boxed{t_1 \cdot t_2 = -1}$$If one end is $(at_1^2, 2at_1)$, the other is $(at_2^2, 2at_2)$ where $t_2 = -\frac{1}{t_1}$
Length of focal chord:
$$\boxed{L = a(t_1 - t_2)^2 = a\left(t_1 + \frac{1}{t_1}\right)^2}$$Harmonic mean of focal chord segments: If focal chord is divided by focus into segments $p$ and $q$:
$$\boxed{\frac{2}{p} + \frac{2}{q} = \frac{1}{a}}$$Or: Semi-latus rectum is the harmonic mean of focal chord segments.
Special Focal Chord - Latus Rectum:
- Perpendicular to axis
- Length = $4a$ (shortest focal chord)
For parabola $y^2 = 8x$, one end of focal chord is at $t = 2$. Find the other end and length of chord.
Solution:
Here $4a = 8$ → $a = 2$
Other end parameter: $t_2 = -\frac{1}{t_1} = -\frac{1}{2}$
Endpoints:
- $(at_1^2, 2at_1) = (2 \cdot 4, 2 \cdot 2 \cdot 2) = (8, 8)$
- $(at_2^2, 2at_2) = (2 \cdot \frac{1}{4}, 2 \cdot 2 \cdot (-\frac{1}{2})) = (\frac{1}{2}, -2)$
Length:
$$L = a(t_1 - t_2)^2 = 2\left(2 - (-\frac{1}{2})\right)^2 = 2\left(\frac{5}{2}\right)^2 = 2 \cdot \frac{25}{4} = \frac{25}{2}$$Chord with Given Midpoint
Equation of chord with midpoint $(x_1, y_1)$ for parabola $y^2 = 4ax$:
$$\boxed{yy_1 - 2a(x + x_1) = y_1^2 - 4ax_1}$$Or more simply:
$$\boxed{T = S_1}$$where $T$ is the tangent equation at $(x_1, y_1)$ and $S_1$ is the value when $(x_1, y_1)$ is substituted in the parabola equation.
Reflection Property of Parabola
The Magic of Satellite Dishes!
Property: Any ray parallel to the axis of the parabola, when reflected, passes through the focus.
Conversely: Any ray from the focus, when reflected, becomes parallel to the axis.
Applications:
- Satellite dishes: Incoming parallel signals (from satellite) reflect and converge at the focus where the receiver is placed
- Car headlights: Bulb at focus emits light that reflects to form parallel beams
- Solar cookers: Parallel sun rays reflect and concentrate at focus, creating intense heat
- Telescopes: Distant star light (parallel rays) converges at focus
Visual Description: Imagine rain falling vertically (parallel to axis) into a parabolic dish - all drops would bounce and meet at the focus!
Memory Tricks & Patterns
Pattern 1: Quick Identification
How to identify a parabola equation:
- Only ONE variable is squared (not both like circle/ellipse)
- No $xy$ term
- Degree is 2
Examples:
- $y^2 = 4ax$ → Parabola ✓
- $x^2 + y^2 = r^2$ → Circle (both squared)
- $xy = c$ → Hyperbola (xy term)
- $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ → Ellipse (both squared, positive)
Pattern 2: The “4a” Pattern
Everything involves 4a:
- Equation: $y^2 = 4ax$
- Latus rectum length: $4a$
- Tangent slope form: $y = mx + \frac{a}{m}$
Memory: “Parabola loves the number 4!”
Pattern 3: Focal Chord Parameter Product
“Focal chord parameters multiply to -1”
$$t_1 \cdot t_2 = -1$$This is THE most important property for JEE Advanced problems!
Common Mistakes to Avoid
Wrong: For $y^2 = 4ax$, focus is at $(4a, 0)$
Correct: Focus is at $(a, 0)$
Remember: Coefficient is $4a$, but distance from vertex to focus is just $a$.
Mnemonic: “Focus is at ‘a’, not ‘4a’”
Common Error: Thinking $y^2 = 4ax$ opens upward
Correct: $y^2 = 4ax$ opens rightward (horizontal parabola)
Rule:
- $y$ squared → horizontal (left/right)
- $x$ squared → vertical (up/down)
For parabola, there are:
- One tangent with a given slope
- Up to three normals with a given slope
Students often think there are multiple tangents like in circles!
Key Difference: Circle is closed curve (tangents on both sides), parabola is open (tangent on one side only).
Wrong: For $y^2 = 4ax$, latus rectum endpoints are $(4a, 4a)$ and $(4a, -4a)$
Correct: Endpoints are $(a, 2a)$ and $(a, -2a)$
Remember:
- x-coordinate = $a$ (focus x-coordinate)
- y-coordinates = $\pm 2a$ (half of latus rectum length $4a$)
Practice Problems
Level 1: Foundation (NCERT Style)
Question: For parabola $y^2 = 20x$, find focus, directrix, vertex, and latus rectum.
Solution:
Compare with $y^2 = 4ax$:
$$4a = 20 \implies a = 5$$- Vertex: $(0, 0)$
- Focus: $(5, 0)$
- Directrix: $x = -5$
- Axis: $y = 0$
- Latus Rectum: $4a = 20$
- LR Endpoints: $(5, 10)$ and $(5, -10)$
Question: Find equation of parabola with vertex at origin, axis along x-axis, and passing through $(3, 6)$.
Solution:
Since axis is along x-axis and vertex at origin, equation is $y^2 = 4ax$.
Point $(3, 6)$ lies on it:
$$6^2 = 4a(3)$$ $$36 = 12a$$ $$a = 3$$ $$\boxed{y^2 = 12x}$$Question: Find coordinates of point on parabola $y^2 = 16x$ with parameter $t = 3$.
Solution:
Here $4a = 16$ → $a = 4$
$$x = at^2 = 4(3)^2 = 36$$ $$y = 2at = 2(4)(3) = 24$$ $$\boxed{(36, 24)}$$Verification: $24^2 = 576 = 16(36)$ ✓
Level 2: JEE Main Type
Question: Find equation of tangent to $y^2 = 12x$ having slope $2$.
Solution:
Here $4a = 12$ → $a = 3$
Using $y = mx + \frac{a}{m}$:
$$y = 2x + \frac{3}{2}$$Multiply by 2:
$$\boxed{2y = 4x + 3 \quad \text{or} \quad 4x - 2y + 3 = 0}$$Point of contact: $\left(\frac{a}{m^2}, \frac{2a}{m}\right) = \left(\frac{3}{4}, 3\right)$
Question: One end of focal chord of parabola $y^2 = 16x$ is $(1, 4)$. Find length of focal chord.
Solution:
Here $4a = 16$ → $a = 4$
Find parameter $t_1$ for point $(1, 4)$:
From $y = 2at$: $4 = 2(4)t$ → $t = \frac{1}{2}$
Other end parameter: $t_2 = -\frac{1}{t_1} = -2$
Length:
$$L = a(t_1 - t_2)^2 = 4\left(\frac{1}{2} - (-2)\right)^2$$ $$L = 4\left(\frac{5}{2}\right)^2 = 4 \times \frac{25}{4} = 25$$Question: Find points of intersection of line $y = 2x + 1$ with parabola $y^2 = 4x$.
Solution:
Substitute $y = 2x + 1$ in $y^2 = 4x$:
$$(2x + 1)^2 = 4x$$ $$4x^2 + 4x + 1 = 4x$$ $$4x^2 + 1 = 0$$ $$x^2 = -\frac{1}{4}$$No real solution!
$$\boxed{\text{Line does not intersect the parabola}}$$(Line is outside the parabola)
Level 3: JEE Advanced Type
Question: Find locus of midpoint of chords of parabola $y^2 = 4ax$ having slope $m$.
Solution:
Let midpoint be $(h, k)$.
Equation of chord with slope $m$ passing through $(h, k)$:
$$y - k = m(x - h)$$ $$y = mx - mh + k \quad \text{...(1)}$$This intersects parabola $y^2 = 4ax$:
$$y^2 = 4ax$$Substitute from (1): $(mx - mh + k)^2 = 4ax$
For $(h, k)$ to be midpoint, use the condition:
Sum of roots formula gives: $y_1 + y_2 = 2k$
From the quadratic in $x$: after expansion and using sum of roots:
The locus equation (replacing $(h,k)$ with $(x,y)$):
$$\boxed{y = \frac{2a}{m}}$$This is a horizontal line!
Interpretation: All midpoints of parallel chords lie on a line parallel to the axis.
Question: Show that locus of point of intersection of perpendicular tangents to parabola $y^2 = 4ax$ is the directrix.
Solution:
Let tangents with slopes $m_1$ and $m_2$ intersect at $(h, k)$.
Tangent equations:
$$y = m_1 x + \frac{a}{m_1}$$ $$y = m_2 x + \frac{a}{m_2}$$Both pass through $(h, k)$:
$$k = m_1 h + \frac{a}{m_1} \quad \text{...(1)}$$ $$k = m_2 h + \frac{a}{m_2} \quad \text{...(2)}$$Perpendicular condition: $m_1 m_2 = -1$
From (1): $km_1 = m_1^2 h + a$
From (2): $km_2 = m_2^2 h + a$
Multiply (1) by $m_1$, (2) by $m_2$ and use $m_1 m_2 = -1$:
After algebra: $h = -a$
Locus: $x = -a$ (the directrix!)
$$\boxed{\text{Locus is the directrix } x = -a}$$Question: A parabolic mirror $y^2 = 4x$ receives light rays parallel to x-axis. Find the point where all reflected rays meet.
Solution:
By the reflection property of parabola: All rays parallel to the axis reflect through the focus.
For $y^2 = 4x$: $4a = 4$ → $a = 1$
Focus: $(1, 0)$
$$\boxed{\text{All rays meet at focus } (1, 0)}$$This is why satellite dishes work! The receiver is placed at the focus.
Quick Revision Box
| Situation | Formula/Approach |
|---|---|
| Standard form (right) | $y^2 = 4ax$ |
| Standard form (up) | $x^2 = 4ay$ |
| Focus distance from vertex | $a$ |
| Latus rectum length | $4a$ |
| Parametric form | $x = at^2$, $y = 2at$ |
| Tangent at $(x_1, y_1)$ | $yy_1 = 2a(x + x_1)$ |
| Tangent at parameter $t$ | $ty = x + at^2$ |
| Tangent with slope $m$ | $y = mx + \frac{a}{m}$ |
| Normal at parameter $t$ | $y + tx = 2at + at^3$ |
| Focal chord property | $t_1 t_2 = -1$ |
| Focal chord length | $a(t_1 - t_2)^2$ |
| Directrix | $x = -a$ (for $y^2=4ax$) |
Cross-Links to Related Topics
Prerequisites:
- Straight Lines - For tangent and normal equations
- Circles - Understanding conic sections
- Quadratic Equations - For solving intersection problems
Related Topics:
- Ellipse - Another conic with two foci
- Hyperbola - Opposite of ellipse
- General Conics - Unified conic approach
- Tangent and Normal to Conics - Tangent techniques
- Chord of Contact - Chord properties
- Definite Integrals - Area under parabola
- Applications of Integrals - Area calculations
Physics Applications:
- Projectile Motion - Parabolic trajectories
- Optics - Parabolic mirrors and reflectors
Other Connections:
- Differentiation - Finding tangent slopes
JEE Exam Strategy
Weightage: 3-4 questions in JEE Main, 2-3 in Advanced
Most Frequently Asked:
- Focal chord properties - especially $t_1 t_2 = -1$ (appears almost every year in Advanced)
- Tangent and normal equations - parametric form preferred
- Locus problems - midpoint of chords, intersection of tangents
- Reflection property - conceptual questions
- Finding equation given conditions - focus, directrix, passing through points
Time-Saving Tricks:
- Always use parametric form for tangent/normal - it’s faster
- For focal chord: Just remember $t_1 t_2 = -1$, calculate everything from this
- Direction recognition: $y^2$ = horizontal, $x^2$ = vertical
- Latus rectum is always $4a$ - don’t calculate, just write!
Common Trap Options:
- Giving focus as $(4a, 0)$ instead of $(a, 0)$
- Confusing tangent equation at point vs with given slope
- Missing the condition $t_1 t_2 = -1$ for focal chord
- Wrong direction (horizontal vs vertical)
Pattern Recognition:
- “Perpendicular tangents” → Think of directrix as locus
- “Focal chord” → Use $t_1 t_2 = -1$
- “Tangent parallel to line” → Use slope form $y = mx + \frac{a}{m}$
- “Reflection” or “satellite dish” → Focus property
Quick Checks:
- After finding equation, verify that focus and directrix are equidistant from vertex
- For focal chord, verify $t_1 t_2 = -1$ before proceeding
- Check if point actually lies on parabola before finding tangent
Teacher’s Summary
- Remember the definition: Parabola is locus of points equidistant from focus and directrix
- Master parametric form $(at^2, 2at)$ - it simplifies 90% of JEE problems
- Focal chord magic: $t_1 t_2 = -1$ is the single most important property
- Direction recognition: $y^2$ = horizontal, $x^2$ = vertical (opposite of intuition!)
- Latus rectum is always $4a$ - the “width” at the focus
- Reflection property explains why satellite dishes, headlights, and solar cookers are parabolic
“The parabola is nature’s focusing curve - master its algebra, and you’ll understand why it appears everywhere from sports to space technology!”
Next Step: Move to Ellipse - the closed curve with two foci, seen in planetary orbits!