IIT JEE Main Maths -Unit 10- Standard equation and parameters (vertex, focus, directrix)- Study Notes-New Syllabus
IIT JEE Main Maths -Unit 10- Standard equation and parameters (vertex, focus, directrix)- Study Notes – New syllabus
IIT JEE Main Maths -Unit 10- Standard equation and parameters (vertex, focus, directrix)- Study Notes -IIT JEE Main Maths – per latest Syllabus.
Key Concepts:
- Parabola: Standard Equation and Parameters
- Latus Rectum of a Parabola
- Parametric Form of Parabola
Parabola: Standard Equation and Parameters
A parabola is the set of all points that are equidistant from a fixed point (focus) and a fixed line (directrix).
Standard equations of parabolas are expressed using the parameter \( a \), which controls the width and shape of the curve.
1. Standard Forms of Parabola
(A) Parabola opening right
\( y^2 = 4ax \)
(B) Parabola opening left
\( y^2 = -4ax \)
(C) Parabola opening up
\( x^2 = 4ay \)
(D) Parabola opening down
\( x^2 = -4ay \)
Here \( a \) is the distance from vertex to focus.
2. Parameters of Parabola
For the standard parabola \( y^2 = 4ax \):
| Feature | Value |
| Vertex | \( (0,0) \) |
| Focus | \( (a,0) \) |
| Directrix | \( x = -a \) |
| Axis | X axis |
| Latus rectum length | \( 4a \) |
| Endpoints of latus rectum | \( (a, 2a),\ (a, -2a) \) |
3. Understanding the Parameter \( a \)
\( a \) is half the distance between the focus and the directrix.
Also:
Larger \( a \) → parabola opens wider Smaller \( a \) → parabola is narrower
4. Standard Forms Together
| Equation | Opens | Vertex | Focus | Directrix |
| \( y^2 = 4ax \) | Right | \( (0,0) \) | \( (a,0) \) | \( x = -a \) |
| \( y^2 = -4ax \) | Left | \( (0,0) \) | \( (-a,0) \) | \( x = a \) |
| \( x^2 = 4ay \) | Upward | \( (0,0) \) | \( (0,a) \) | \( y = -a \) |
| \( x^2 = -4ay \) | Downward | \( (0,0) \) | \( (0,-a) \) | \( y = a \) |
Example
For parabola \( y^2 = 12x \), find its focus and directrix.
▶️ Answer / Explanation
Given: \( 4a = 12 \Rightarrow a = 3 \)
Focus: \( (3,0) \)
Directrix: \( x = -3 \)
Example
Find the equation of a parabola with focus \( (2,0) \) and directrix \( x = -2 \).
▶️ Answer / Explanation
Focus distance from vertex = a
\( a = 2 \)
Opens right → equation is:
\( y^2 = 4ax = 8x \)
Example
Find the equation of a parabola whose vertex is at origin and focus is \( (0,-4) \).
▶️ Answer / Explanation
Vertex at (0,0)
Focus at (0,-4) → parabola opens downward.
So use form:
\( x^2 = -4ay \)
Here focus is at (0,-a) → \( -a = -4 \Rightarrow a = 4 \).
Equation:
\( x^2 = -16y \)
Latus Rectum of a Parabola
The latus rectum is a very important line segment in parabola questions. It is a chord of the parabola:
- passing through focus
- perpendicular to the axis of the parabola
- with fixed length depending only on parameter \( a \)
1. Latus Rectum for Standard Parabolas
For the parabola
\( y^2 = 4ax \)
- Focus: \( (a,0) \)
- Axis: X-axis
- Latus rectum: perpendicular to axis → vertical line through focus
Equation of Latus Rectum:
\( x = a \)
Length of Latus Rectum:
\( 4a \)
Endpoints of Latus Rectum:
\( (a, 2a),\ (a, -2a) \)
2. Latus Rectum for All Four Standard Parabolas
| Parabola | Focus | Latus Rectum Line | Endpoints | Length |
| \( y^2 = 4ax \) | \( (a,0) \) | \( x = a \) | \( (a,2a),\ (a,-2a) \) | \( 4a \) |
| \( y^2 = -4ax \) | \( (-a,0) \) | \( x = -a \) | \( (-a,2a),\ (-a,-2a) \) | \( 4a \) |
| \( x^2 = 4ay \) | \( (0,a) \) | \( y = a \) | \( (2a,a),\ (-2a,a) \) | \( 4a \) |
| \( x^2 = -4ay \) | \( (0,-a) \) | \( y = -a \) | \( (2a,-a),\ (-2a,-a) \) | \( 4a \) |
3. How to Find Latus Rectum for Translated Parabolas
If parabola is shifted, example:
\( (y – k)^2 = 4a(x – h) \)
- Focus moves to \( (a + h,\ k) \)
- Latus rectum is vertical line through focus → \( x = a + h \)
- Endpoints become: \( (a + h,\ k + 2a),\ (a + h,\ k – 2a) \)
Example
Find the latus rectum length and endpoints for the parabola \( y^2 = 8x \).
▶️ Answer / Explanation
\( 8 = 4a \Rightarrow a = 2 \)
Length: \( 4a = 8 \)
Endpoints: \( (2, 4),\ (2, -4) \)
Example
Find the equation of latus rectum of parabola \( x^2 = 12y \).
▶️ Answer / Explanation
\( 12 = 4a \Rightarrow a = 3 \)
Parabola opens upward → latus rectum is horizontal line.
Equation:
\( y = a = 3 \)
Example
For the translated parabola \( (y – 1)^2 = 4(x + 2) \), find the endpoints of the latus rectum.
▶️ Answer / Explanation
Standard form:
\( (y – k)^2 = 4a(x – h) \)
Compare:
\( k = 1,\quad h = -2,\quad 4a = 4 \Rightarrow a = 1 \)
Focus: \( (a + h,\ k) = (1 – 2,\ 1) = (-1,\ 1) \)
Latus rectum is vertical line → through focus:
\( x = -1 \)
Endpoints:
\( (-1,\ 1 + 2a) = (-1,\ 3) \)
\( (-1,\ 1 – 2a) = (-1,\ -1) \)
Answer: endpoints are \( (-1,3) \) and \( (-1,-1) \)
Parametric Form of Parabola
Parametric coordinates are extremely important because many tangent, normal, chord, and locus questions in JEE become simpler using parameter \( t \).
For any standard parabola, we assign a parameter to generate every point on the parabola.
1. Parametric Coordinates for Standard Parabola
(A) For parabola \( y^2 = 4ax \)
The parametric coordinates of any point on the parabola are:
\( (x,\ y) = (at^2,\ 2at) \)
As \( t \) takes all real values, the point moves along the entire parabola.
2. Why Parametric Form Works
Substitute \( x = at^2,\ y = 2at \) into parabola:
Left side: \( y^2 = (2at)^2 = 4a^2 t^2 \)
Right side: \( 4ax = 4a(at^2) = 4a^2 t^2 \)
Hence identity holds → every such point lies on the parabola.
3. Parametric Forms of All Standard Parabolas
| Parabola | Parametric Point |
| \( y^2 = 4ax \) | \( (at^2,\ 2at) \) |
| \( y^2 = -4ax \) | \( (-at^2,\ 2at) \) |
| \( x^2 = 4ay \) | \( (2at,\ at^2) \) |
| \( x^2 = -4ay \) | \( (2at,\ -at^2) \) |
4. Parametric Form for Translated Parabolas
For parabola:
\((y – k)^2 = 4a(x – h)\)
The parametric point is:
\( (x,\ y) = (h + at^2,\ k + 2at) \)
5. Very Important Uses of Parametric Form
- Equation of tangent becomes very easy
- Equation of normal becomes simple
- Finding chord properties becomes simpler
- Distances from focus are easier
- Angle and slope calculations become straightforward
6. Special JEE Note
Parameter \( t \) represents the slope of tangent at the parametric point for the parabola:
\( y^2 = 4ax \Rightarrow m = \dfrac{y}{x} = \dfrac{2at}{at^2} = \dfrac{2}{t} \)
This is used for tangent, normal, and locus problems
Example
Find the parametric coordinates of the point on parabola \( y^2 = 16x \) for parameter \( t = 3 \).
▶️ Answer / Explanation
Given: \( 4a = 16 \Rightarrow a = 4 \)
Parametric point:
\( (x,y) = (at^2,\ 2at) \)
\( x = 4(9) = 36 \)
\( y = 2(4)(3) = 24 \)
Answer: \( (36, 24) \)
Example
Find the parametric equation of the parabola \((y – 2)^2 = 12(x + 1)\).
▶️ Answer / Explanation
\( 4a = 12 \Rightarrow a = 3 \)
Standard equivalent:
\((y – 2)^2 = 4a(x + 1)\)
Thus parametric point is:
\( (x,y) = (-1 + 3t^2,\ 2 + 6t) \)
Answer: \( (-1 + 3t^2,\ 2 + 6t) \)
Example
The point \( (9,12) \) lies on a parabola of the form \( y^2 = 4ax \). Find the parameter \( t \) corresponding to this point.
▶️ Answer / Explanation
Parametric point:
\( x = at^2,\ y = 2at \)
From y equation:
\( 12 = 2at \Rightarrow at = 6 \)
From x equation:
\( 9 = at^2 \Rightarrow a t^2 = 9 \)
Divide the two equations:
\( \dfrac{at^2}{at} = \dfrac{9}{6} \)
\( t = \dfrac{3}{2} \)
Answer: \( t = \dfrac{3}{2} \)
Notes and Study Materials
- Concepts of Parabola
- Parabola Master File
- Parabola Revision Notes
- Parabola Formulae
- Conic Section Formulae
- Parabola Reference Book
- Parabola Past Many Years Questions and Answer
Examples and Exercise
IIT JEE (Main) Mathematics ,”Parabola” Notes ,Test Papers, Sample Papers, Past Years Papers , NCERT , S. L. Loney and Hall & Knight Solutions and Help from Ex- IITian
About this unit
Parabola
IITian Academy Notes for IIT JEE (Main) Mathematics – Parabola
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