JEE Maths — Conic Sections Complete Chapter Guide

Chapter Overview & Weightage

Conic Sections is one of the most consistently weighted chapters in JEE Maths — it rewards students who understand the geometry, not just the formulas. Every year, you’ll see 2-3 questions in JEE Main and at least 1-2 in JEE Advanced, often in the form of paragraph-based or multi-correct problems.

Weightage pattern: 8-10% of JEE Main Maths (roughly 3-4 marks per session). In JEE Advanced, conic questions appear in integer type or comprehension format — typically 1 question worth 3-4 marks. The chapter combines well with Straight Lines, so questions often test both simultaneously.

Year	JEE Main Questions	Marks	JEE Advanced
2024	3	12	1 (comprehension)
2023	3	12	2
2022	2-3	8-12	1-2
2021	3	12	1
2020	2	8	1

Parabola gets the most love in JEE Main. Ellipse and hyperbola tend to appear in JEE Advanced with more depth — locus problems, chord of contact, and combined properties.

Key Concepts You Must Know

Prioritised by exam frequency:

Parabola (highest frequency)

Standard form $y^2 = 4ax$ and its four orientations
Focal chord: if one end is $(at^2, 2at)$ , the other end is $\left(\frac{a}{t^2}, -\frac{2a}{t}\right)$
Length of focal chord with parameter $t$ : $a\left(t + \frac{1}{t}\right)^2$
Condition for a line $y = mx + c$ to be tangent: $c = \frac{a}{m}$
Foot of normal, three normals from an external point

Ellipse (medium frequency)

Standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$ , relation $b^2 = a^2(1-e^2)$
Sum of focal distances: $PS + PS' = 2a$ (this property appears in locus questions constantly)
Auxiliary circle, eccentric angle, parametric form $(a\cos\theta, b\sin\theta)$
Director circle: $x^2 + y^2 = a^2 + b^2$

Hyperbola (medium-high in Advanced)

Standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ , asymptotes $y = \pm\frac{b}{a}x$
Difference of focal distances: $|PS - PS'| = 2a$
Rectangular hyperbola $xy = c^2$ : parametric form $(ct, c/t)$ — appears often in JEE Advanced
Conjugate hyperbola and its relation to asymptotes

Chord, Tangent, Normal — for all conics

Chord of contact from external point $(x_1, y_1)$ to ellipse: $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$
Chord with midpoint $(h, k)$ : use $T = S_1$ (the single most useful shortcut in this chapter)
Normal at parametric point — slope of normal is always negative reciprocal of slope of tangent

Important Formulas

Parametric point: $(at^2, 2at)$

Tangent at $(at^2, 2at)$ : $ty = x + at^2$

Normal at $(at^2, 2at)$ : $y + tx = 2at + at^3$

Condition for tangency ( $y = mx + c$ ): $c = \dfrac{a}{m}$ , tangent is $y = mx + \dfrac{a}{m}$

Length of latus rectum: $4a$

Focal chord length: $a\left(t + \dfrac{1}{t}\right)^2$

Minimum focal chord length: $4a$ (latus rectum itself)

When to use: Parametric form is your default for parabola — almost every tangent/normal/chord problem becomes cleaner with it. Use Cartesian form only when the question gives you a specific point.

Eccentricity: $e = \sqrt{1 - b^2/a^2}$ , foci at $(\pm ae, 0)$

Parametric point: $(a\cos\theta, b\sin\theta)$

Tangent at $(a\cos\theta, b\sin\theta)$ : $\dfrac{x\cos\theta}{a} + \dfrac{y\sin\theta}{b} = 1$

Chord with midpoint $(h,k)$ : $\dfrac{hx}{a^2} + \dfrac{ky}{b^2} = \dfrac{h^2}{a^2} + \dfrac{k^2}{b^2}$

Director circle: $x^2 + y^2 = a^2 + b^2$

Sum of focal radii: $r_1 + r_2 = 2a$

When to use: The $T = S_1$ formula (chord with given midpoint) is the fastest approach when a question gives you midpoint and asks for the chord equation. Memorise this pattern — it saves 3-4 minutes.

Eccentricity: $e = \sqrt{1 + b^2/a^2} > 1$ , foci at $(\pm ae, 0)$

Asymptotes: $\dfrac{x}{a} \pm \dfrac{y}{b} = 0$ , i.e., $bx \pm ay = 0$

Rectangular hyperbola $xy = c^2$ : parametric $(ct, c/t)$ , tangent $\dfrac{x}{t} + yt = 2c$

Difference of focal radii: $|r_1 - r_2| = 2a$

When to use: For rectangular hyperbola $xy = c^2$ , always use parametric form. Questions on chord of contact, normals, and locus become mechanical once you’re comfortable with $(ct, c/t)$ .

Solved Previous Year Questions

PYQ 1 — JEE Main 2024 (January, Shift 2)

Q. The line $y = mx + 1$ is tangent to the parabola $y^2 = 4x$ . Find $m$ .

Solution:

For $y^2 = 4x$ , we have $a = 1$ . The condition for $y = mx + c$ to be tangent is $c = \dfrac{a}{m} = \dfrac{1}{m}$ .

Here $c = 1$ , so $1 = \dfrac{1}{m}$ , giving $m = 1$ .

Answer: $m = 1$

Students often forget the tangency condition and instead try substituting and setting discriminant $= 0$ . That works but takes twice as long. The formula $c = a/m$ is your 10-second shortcut — memorise it cold.

PYQ 2 — JEE Main 2023 (April, Shift 1)

Q. If the chord joining the points $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$ on the parabola $y^2 = 4ax$ passes through the focus, show that $t_1 t_2 = -1$ .

Solution:

The focus of $y^2 = 4ax$ is $(a, 0)$ .

Slope of $PQ$ :

m = \frac{2at_1 - 2at_2}{at_1^2 - at_2^2} = \frac{2(t_1 - t_2)}{(t_1-t_2)(t_1+t_2)} = \frac{2}{t_1 + t_2}

Equation of chord $PQ$ using point $P$ :

y - 2at_1 = \frac{2}{t_1+t_2}(x - at_1^2)

Substituting the focus $(a, 0)$ :

0 - 2at_1 = \frac{2}{t_1+t_2}(a - at_1^2)

-2at_1(t_1+t_2) = 2a(1 - t_1^2)

-t_1(t_1+t_2) = 1 - t_1^2

-t_1^2 - t_1 t_2 = 1 - t_1^2

-t_1 t_2 = 1

\boxed{t_1 t_2 = -1}

This result — $t_1 t_2 = -1$ for a focal chord of a parabola — is so frequently tested that you should remember it as a property, not derive it every time. If the problem says “focal chord”, immediately write $t_1 t_2 = -1$ .

PYQ 3 — JEE Advanced 2022 (Paper 1)

Q. Let the eccentricity of the hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ be $\dfrac{5}{4}$ . If the equation of the normal at the point $\left(\dfrac{8}{\sqrt{5}}, \dfrac{12}{5}\right)$ is $8x + \beta y = \lambda$ , find $\lambda - \beta$ .

Solution:

Given $e = \dfrac{5}{4}$ , so $e^2 = \dfrac{25}{16}$ .

Using $e^2 = 1 + \dfrac{b^2}{a^2}$ :

\frac{b^2}{a^2} = \frac{25}{16} - 1 = \frac{9}{16}

Let $a^2 = 16k$ and $b^2 = 9k$ . Verify the given point lies on the hyperbola:

\frac{(8/\sqrt{5})^2}{16k} - \frac{(12/5)^2}{9k} = \frac{64/5}{16k} - \frac{144/25}{9k} = \frac{4}{5k} - \frac{16}{25k} = \frac{20-16}{25k} = \frac{4}{25k}

For this to equal 1: $k = \dfrac{4}{25}$ , so $a^2 = \dfrac{64}{25}$ , $b^2 = \dfrac{36}{25}$ .

The normal to $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ at $(x_1, y_1)$ :

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

At $\left(\dfrac{8}{\sqrt{5}}, \dfrac{12}{5}\right)$ :

\frac{(64/25)x}{8/\sqrt{5}} + \frac{(36/25)y}{12/5} = \frac{64}{25} + \frac{36}{25} = 4

\frac{8x}{5\sqrt{5}} + \frac{3y}{5} = 4

Multiply by $5\sqrt{5}$ : $8x + 3\sqrt{5}\,y = 20\sqrt{5}$

So $\beta = 3\sqrt{5}$ , $\lambda = 20\sqrt{5}$ , giving $\lambda - \beta = 17\sqrt{5}$ .

Difficulty Distribution

For JEE Main, the chapter typically breaks as:

Difficulty	% of Questions	What to Expect
Easy	30%	Standard tangent/normal, basic focal properties
Medium	50%	Locus problems, chord with given midpoint, two-conic combined
Hard	20%	Family of conics, three normals from a point, asymptote properties

For JEE Advanced, expect mostly medium-to-hard — often involving 2-3 conics interacting, or a locus that simplifies to a conic you need to identify.

In JEE Main 2024 and 2023, parabola accounted for roughly 60% of all conic questions. If time is tight, prioritise parabola completely before moving to ellipse, then hyperbola last.

Expert Strategy

Week 1 of preparation: Lock down parabola — standard form, parametric equations, tangent and normal conditions, focal chord property. Solve 20-25 parabola questions from PYQs only.

Week 2: Ellipse — focus on the $T = S_1$ midpoint chord trick and the auxiliary circle concept. These appear in Advanced-style questions and take time to internalise.

Week 3: Hyperbola and rectangular hyperbola. The rectangular hyperbola $xy = c^2$ is disproportionately high-value for the effort it takes — 2 days of focused practice makes it mechanical.

The locus shortcut: When a question asks “find the locus of…”, assume it will be a conic. Write the parametric coordinates of your point, eliminate the parameter, and simplify. If your algebra is giving something messy, you’ve likely made a sign error — backtrack immediately rather than pushing forward.

For PYQ practice: Solve JEE Main 2019-2024 questions chapter-wise. You’ll notice the same 5-6 question types recycling with different numbers. Once you’ve seen the pattern, scoring full marks in this chapter is realistic.

Time allocation in exam: A medium conic question should take you 2-3 minutes max. If you’re past 4 minutes, mark it and return — these questions can become time sinks that wreck your overall score.

Common Traps

Trap 1 — Orientation error: $y^2 = 4ax$ opens right, $y^2 = -4ax$ opens left, $x^2 = 4ay$ opens up, $x^2 = -4ay$ opens down. Students mix up the focus coordinates, especially for the vertically oriented parabola where the focus is $(0, a)$ not $(a, 0)$ . Always sketch the orientation before writing focus/directrix.

Trap 2 — Ellipse condition $a > b$ : The formula $b^2 = a^2(1-e^2)$ assumes $a > b$ . If the question gives you $\frac{x^2}{9} + \frac{y^2}{16} = 1$ , here $b > a$ and the major axis is along the $y$ -axis. The foci are at $(0, \pm be)$ , not $(\pm ae, 0)$ . This trips up a significant number of students even at the JEE level.

Trap 3 — Normal vs Tangent slopes: The slope of the tangent to $y^2 = 4ax$ at $(at^2, 2at)$ is $\frac{1}{t}$ . The normal has slope $-t$ . Students regularly flip these. The mnemonic: tangent has $t$ in denominator (small slope for large $t$ ), normal has $-t$ in numerator (large slope for large $t$ ).

Trap 4 — Focal chord minimum length: The minimum length of a focal chord of $y^2 = 4ax$ is $4a$ (the latus rectum). Many students forget to check whether a minimum is asked and pick an arbitrary chord. If the question asks “minimum length of focal chord”, the answer is always the latus rectum length.

Trap 5 — Asymptote vs directrix confusion in hyperbola: The asymptotes of $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are $bx \pm ay = 0$ . These are NOT the directrices. The directrices are $x = \pm \frac{a}{e}$ . Exam setters deliberately structure questions so that confusing these two gives a clean but wrong answer.