Is Parabola part of the EAPCET Mathematics syllabus?

Yes. Parabola is taught in Intermediate second year · Paper IIB Mathematics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.

How many Mathematics questions are there in EAPCET?

80 of the 160 questions in the Engineering paper are Mathematics, spread across 37 chapters.

What mistakes do students make in Parabola?

Taking a = 12 for y² = 12x; compare with y² = 4ax to get a = 3. Using focus (a, 0) for x² = 4ay; for a parabola opening upward the focus is (0, a) and the directrix is y = −a. Writing the tangency condition for y² = 4ax as c = am; the correct condition is c = a/m. Using t₁t₂ = 1 for a focal chord; the ends of a focal chord satisfy t₁t₂ = −1.

EAPCET Mathematics · Intermediate second year · Paper IIB

Parabola

Parabola covers the standard parabola, its focus, directrix and latus rectum, and its tangents and normals, using the parametric form.

Subject: Mathematics
Taught in: Intermediate second year · Paper IIB
Mathematics in the paper: 80 of 160 questions

Key formulas and results

$y^{2} = 4ax$ : focus $(a, 0)$ , directrix $x = -a$ , latus rectum length $4a$
Parametric point: $(at^{2}, 2at)$
Tangent at $t$ : $ty = x + at^{2}$
$y = mx + c$ touches $y^{2} = 4ax$ iff $c = \dfrac{a}{m}$
Normal at $t$ : $y + tx = 2at + at^{3}$
Ends of a focal chord satisfy $t_{1}t_{2} = -1$
Length of the focal chord of $y^{2} = 4ax$ inclined at $\alpha$ to the axis: $4a\operatorname{cosec}^{2}\alpha$
Length of the focal chord of $y^{2} = 4ax$ with one end at $(at^{2}, 2at)$ : $a\left(t + \dfrac{1}{t}\right)^{2}$
Point of intersection of the tangents at $t_{1}$ and $t_{2}$ on $y^{2} = 4ax$ : $(at_{1}t_{2}, a(t_{1} + t_{2}))$
Normal to $y^{2} = 4ax$ with slope $m$ : $y = mx - 2am - am^{3}$
If the normal at $t_{1}$ meets $y^{2} = 4ax$ again at $t_{2}$ , then $t_{2} = -t_{1} - \dfrac{2}{t_{1}}$
$y = mx + c$ touches $x^{2} = 4ay$ iff $c = -am^{2}$ ; $y = mx + \dfrac{a}{m}$ touches $y^{2} = 4ax$ at $\left(\dfrac{a}{m^{2}}, \dfrac{2a}{m}\right)$
Area of the triangle formed by the tangents from $(x_{1}, y_{1})$ to $y^{2} = 4ax$ and their chord of contact: $\dfrac{(y_{1}^{2} - 4ax_{1})^{3/2}}{2a}$
Length of the chord of contact from $(x_{1}, y_{1})$ to $y^{2} = 4ax$ : $\dfrac{\sqrt{y_{1}^{2} - 4ax_{1}}\,\sqrt{y_{1}^{2} + 4a^{2}}}{a}$

What this chapter covers

Standard forms
Focus, directrix, vertex and latus rectum
Parametric form
Tangents and normals
Focal chords

How it is usually asked

Find the focus, directrix or latus rectum
Write tangent and normal equations
Use focal chord properties

Understanding Parabola

A parabola is the set of points equidistant from a fixed point, the focus, and a fixed line, the directrix, so its eccentricity is 1. In the standard form y² = 4ax the vertex is the origin, the focus is (a, 0), the directrix is x = −a and the axis is the x-axis. The latus rectum is the chord through the focus perpendicular to the axis, with length 4a. The other three standard forms open left, up or down.

Every point on y² = 4ax can be written as (at², 2at) for a single parameter t. This parametric form makes the formulas short: the tangent at t is ty = x + at², the normal at t is y + tx = 2at + at³, and the chord joining t₁ and t₂ is a focal chord exactly when t₁t₂ = −1. Using t turns many two-variable problems into one-variable algebra.

In slope form, y = mx + a/m is a tangent to y² = 4ax for every non-zero m, so a line y = mx + c touches the parabola when c = a/m. EAPCET questions ask you to find the focus, directrix or latus rectum of a given parabola, write tangents and normals, or use focal chord properties, and the first step is always reading off a correctly.

Worked example

EAPCET-style question

One end of a focal chord of the parabola $y^{2} = 12x$ is $(27, 18)$ . The other end is

(1) $\left(\tfrac{1}{3}, -2\right)$
(2) $\left(\tfrac{1}{3}, 2\right)$
(3) $(3, -6)$
(4) $(27, -18)$

Solution

$y^{2} = 12x$ gives $4a = 12$ , so $a = 3$ and points are $(3t^{2}, 6t)$ .
$(27, 18)$ corresponds to $t_{1} = 3$ , since $3(9) = 27$ and $6(3) = 18$ .
For a focal chord $t_{1}t_{2} = -1$ , so $t_{2} = -\tfrac{1}{3}$ .
The other end is $\left(3 \times \tfrac{1}{9}, 6 \times \left(-\tfrac{1}{3}\right)\right) = \left(\tfrac{1}{3}, -2\right)$ , which satisfies $y^{2} = 4 = 12 \times \tfrac{1}{3}$ .

Answer: $\left(\tfrac{1}{3}, -2\right)$

Shortcuts and results to remember

The latus rectum is the shortest focal chord, so every focal chord has length at least $4a$ .
If a focal chord is cut by the focus into segments $l_{1}$ and $l_{2}$ , the latus rectum equals $\dfrac{4l_{1}l_{2}}{l_{1} + l_{2}}$ (the semi-latus rectum is their harmonic mean).
Three real normals can be drawn from $(h, k)$ to $y^{2} = 4ax$ only if $h > 2a$ ; their slopes satisfy $m_{1} + m_{2} + m_{3} = 0$ and $m_{1}m_{2}m_{3} = -\dfrac{k}{a}$ .
The image of the focus in any tangent lies on the directrix.
Vertex to focus = vertex to directrix $= a$ , and focus to directrix $= 2a$ .

Common mistakes to avoid

Taking a = 12 for y² = 12x; compare with y² = 4ax to get a = 3.
Using focus (a, 0) for x² = 4ay; for a parabola opening upward the focus is (0, a) and the directrix is y = −a.
Writing the tangency condition for y² = 4ax as c = am; the correct condition is c = a/m.
Using t₁t₂ = 1 for a focal chord; the ends of a focal chord satisfy t₁t₂ = −1.

How to prepare it

Work in the parametric form (at², 2at) — it turns most parabola problems into one-variable algebra.

Once the formulas feel familiar, test them the way the exam does: practise Parabola questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Mathematics chapters

See every chapter in the EAPCET Mathematics syllabus.

Practise Parabola chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Parabola part of the EAPCET Mathematics syllabus?: Yes. Parabola is taught in Intermediate second year · Paper IIB Mathematics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Parabola for EAPCET?: Work in the parametric form (at², 2at) — it turns most parabola problems into one-variable algebra.
How many Mathematics questions are there in EAPCET?: 80 of the 160 questions in the Engineering paper are Mathematics, spread across 37 chapters.
What mistakes do students make in Parabola?: Taking a = 12 for y² = 12x; compare with y² = 4ax to get a = 3. Using focus (a, 0) for x² = 4ay; for a parabola opening upward the focus is (0, a) and the directrix is y = −a. Writing the tangency condition for y² = 4ax as c = am; the correct condition is c = a/m. Using t₁t₂ = 1 for a focal chord; the ends of a focal chord satisfy t₁t₂ = −1.