Comprehensive guide and practice questions for CBSE Class 11 Applied Mathematics curriculum (2026-27).
1. Fundamental Conic Definition of a Parabola
In analytical geometry, a parabola is defined as the locus of a point $P(x, y)$ that moves in a plane such that its distance from a fixed point $S$ (called the focus) is strictly equal to its perpendicular distance from a fixed straight line $L$ (called the directrix).
Conic Ratio and Eccentricity Condition
$$\frac{SP}{PM} = e = 1 \implies SP = PM$$
Where:
$S$ is the fixed focus.
$M$ is the foot of the perpendicular drawn from moving point $P$ to the directrix line $L$.
$e = 1$ is the eccentricity characteristic of all parabolas.
Key Structural Terminology:
Axis of the Parabola: The straight line passing through the focus perpendicular to the directrix. The parabola is completely symmetrical about its axis.
Vertex ($V$): The point of intersection of the parabola with its axis. The vertex is the exact midpoint between the focus $S$ and the directrix.
Focal Chord: Any line segment passing through the focus and having both its endpoints on the parabola.
Latus Rectum: The specific focal chord drawn perpendicular to the axis of the parabola.
Double Ordinate: Any chord drawn perpendicular to the axis of symmetry.
2. Four Standard Orientations of a Parabola
When the vertex is located at the origin $O(0, 0)$ and the axis coincides with one of the coordinate axes, parabolas assume four canonical equations depending on their orientation:
Characteristic
Rightward ($y^2 = 4ax$)
Leftward ($y^2 = -4ax$)
Upward ($x^2 = 4ay$)
Downward ($x^2 = -4ay$)
Orientation
Opens towards $+x$
Opens towards $-x$
Opens towards $+y$
Opens towards $-y$
Focus ($S$)
$(a, 0)$
$(-a, 0)$
$(0, a)$
$(0, -a)$
Directrix Equation
$x = -a$ ($x + a = 0$)
$x = a$ ($x - a = 0$)
$y = -a$ ($y + a = 0$)
$y = a$ ($y - a = 0$)
Axis of Symmetry
$y = 0$ (x-axis)
$y = 0$ (x-axis)
$x = 0$ (y-axis)
$x = 0$ (y-axis)
Vertex ($V$)
$(0, 0)$
$(0, 0)$
$(0, 0)$
$(0, 0)$
Length of Latus Rectum
$4a$
$4a$
$4a$
$4a$
Latus Rectum Endpoints
$(a, \pm 2a)$
$(-a, \pm 2a)$
$(\pm 2a, a)$
$(\pm 2a, -a)$
Focal Distance of $P(x_1, y_1)$
$x_1 + a$
$a - x_1$
$y_1 + a$
$a - y_1$
Parameter Convention ($a > 0$): The constant $a$ represents the physical geometric distance from the vertex to the focus, and identically from the vertex to the directrix. Hence, $a$ is always positive. The total distance between the focus and directrix is $2a$.
3. Step-by-Step Worked Examples
Example 1: Parameter Identification from Standard Equation
Problem: Find the coordinates of the focus, equation of the directrix, axis of symmetry, and length of the latus rectum for the parabola $x^2 = -12y$.
Solution:
Comparing with standard form $x^2 = -4ay$:
$$4a = 12 \implies a = 3$$
Since the equation has $x^2$ and a negative linear term in $y$, this is a downward-opening parabola symmetric about the y-axis.
Focus: $(0, -a) = (0, -3)$.
Directrix: $y = a \implies y = 3$ (or $y - 3 = 0$).
Axis of Symmetry: $x = 0$ (the y-axis).
Length of Latus Rectum: $4a = 12\text{ units}$.
Example 2: Determining Equation from Vertex and a Boundary Point
Problem: A parabola has its vertex at the origin $(0, 0)$, its axis along the x-axis, and its curve passes through the coordinate point $(3, -6)$. Find its Cartesian equation and focus.
Solution:
Since the axis is along the x-axis, the equation is of the form $y^2 = 4ax$ or $y^2 = -4ax$.
The point $(3, -6)$ has a positive x-coordinate ($x = 3 > 0$). Hence, the parabola opens towards the positive x-axis:
$$y^2 = 4ax$$
Substitute $(3, -6)$ into the equation:
$$(-6)^2 = 4a(3) \implies 36 = 12a \implies a = 3$$
Equation of the parabola:
$$y^2 = 4(3)x \implies y^2 = 12x$$
Focus is located at $(a, 0) = (3, 0)$.
Example 3: Applied Real-World Case Study — Parabolic Solar Thermal Collector
Problem: An industrial thermal engineering company designs a parabolic trough collector to concentrate incident solar radiation onto a central fluid pipe. The parabolic reflector has an aperture diameter of $4\text{ meters}$ across its rim and a central depth of $1\text{ meter}$.
(a) Determine the Cartesian equation of the parabolic profile placing the vertex at the origin.
(b) At what exact distance from the vertex should the heat-absorbing pipe (focus) be mounted to maximize concentrated thermal absorption?
Solution:
Model Formulation:
Let the vertex be at $V(0, 0)$ and orient the parabola opening upward along the y-axis: $x^2 = 4ay$.
The aperture width is $4\text{ m}$, spanning horizontally from $x = -2$ to $x = +2$.
At the outer rim ($x = 2$), the depth is $y = 1\text{ m}$. Thus, point $(2, 1)$ lies on the parabola.
Placement of Absorber Pipe:
The reflective property of parabolas ensures that all incoming parallel rays of sunlight reflect directly through the focus $S(0, a) = (0, 1)$.
Therefore, the heat absorber pipe must be positioned exactly $1\text{ meter}$ above the vertex (level with the outer rim plane).
Progress Check & Skill Assessment
Solve these structured exercises to test your mastery of focal geometry, directrix equations, and applied parabolic designs.
Tier 1: Foundations & Parameter Derivation
Problem 1: Find the focus, vertex, directrix, and latus rectum of the parabola $y^2 = 20x$.
View Complete Solution
Comparing with $y^2 = 4ax$: $4a = 20 \implies a = 5$.
Vertex: $(0, 0)$
Focus: $(a, 0) = (5, 0)$
Directrix: $x = -a \implies x = -5$ (or $x + 5 = 0$)
Length of Latus Rectum: $4a = 20\text{ units}$
Problem 2: Find the equation of the parabola with focus at $(0, -4)$ and directrix $y = 4$.
View Complete Solution
The focus lies on the negative y-axis at $(0, -a) = (0, -4) \implies a = 4$.
The directrix is $y = a = 4$.
This corresponds to a downward-opening vertical parabola: $x^2 = -4ay$.
Equation: $x^2 = -4(4)y \implies x^2 = -16y$.
Tier 2: Geometric & Focal Distance Analysis
Problem 3: Find the coordinates of a point on the parabola $y^2 = 8x$ whose focal distance is $6\text{ units}$.
View Complete Solution
$4a = 8 \implies a = 2$.
For $y^2 = 4ax$, the focal distance of any point $P(x, y)$ is given by $x + a$.
Endpoints are $(-6, 3)$ and $(6, 3)$. Distance between them $= 6 - (-6) = 12 = 4a$.
Tier 3: Applied Structural & Engineering Models
Problem 5: A suspension bridge cable hangs in a parabolic arc between two towers $120\text{ m}$ apart. The top of each tower is $30\text{ m}$ above the road deck, and the lowest point of the cable is $6\text{ m}$ above the road deck. Taking the origin at the vertex (lowest point of the cable), determine the height of the cable at a horizontal distance of $30\text{ m}$ from the center.
View Complete Solution
Let vertex be $V(0, 0)$ and parabola equation be $x^2 = 4ay$.
The span is $120\text{ m}$, so towers are at $x = \pm 60\text{ m}$.
4. An arch is in the form of a parabola with vertical axis. The arch is $10\text{ m}$ high and $5\text{ m}$ wide at the base. How wide is it at a height of $2\text{ m}$ from the vertex (top)?
Show Solution
Place vertex at origin $(0, 0)$ with axis vertical downwards: $x^2 = -4ay$.
At the base, $y = -10\text{ m}$, width is $5\text{ m}$ ($x = \pm 2.5$).
At $2\text{ m}$ below vertex ($y = -2$): $x^2 = 0.625(2) = 1.25 \implies x = \sqrt{1.25} \approx 1.118\text{ m}$.
Total width $= 2x = 2(1.118) \approx 2.24\text{ meters}$.
Quick Reference & High-Yield Summary
Equation
Axis
Focus
Directrix
Latus Rectum
Focal Length ($SP$)
$y^2 = 4ax$
$y = 0$ (x-axis)
$(a, 0)$
$x = -a$
$4a$
$x_1 + a$
$y^2 = -4ax$
$y = 0$ (x-axis)
$(-a, 0)$
$x = a$
$4a$
$a - x_1$
$x^2 = 4ay$
$x = 0$ (y-axis)
$(0, a)$
$y = -a$
$4a$
$y_1 + a$
$x^2 = -4ay$
$x = 0$ (y-axis)
$(0, -a)$
$y = a$
$4a$
$a - y_1$
High-Yield Exam Reminders:
Eccentricity: For any parabola, $e = 1$ unconditionally.
Vertex is Midpoint: The vertex $(0, 0)$ always bisects the distance between the focus and the foot of the directrix on the axis.
Optical Property: Any ray originating from the focus and striking the interior reflects parallel to the axis of symmetry (used in automotive headlights and flashlights). Conversely, any ray parallel to the axis reflects through the focus (used in satellite dishes and radio telescopes).
10-Question Diagnostic Assessment: Parabola
Attempt all 10 diagnostic questions covering standard equations, focal geometry, directrices, latus rectum properties, and applied physical systems. Click Submit Diagnostic Mock to evaluate your answers with detailed step-by-step solutions.