Conic Sections — Complete Guide with Solved Examples

What Are Conic Sections — And Why Do They Matter?

Take a double cone — two ice cream cones joined tip-to-tip — and slice through it with a flat plane. Depending on the angle of the cut, you get four different curves: a circle, an ellipse, a parabola, or a hyperbola. That’s the entire story of conic sections in one image.

These curves show up everywhere: satellite dishes are parabolic, planetary orbits are elliptical, and cooling tower silhouettes are hyperbolic. But for us right now, the bigger deal is that conic sections carry 2–3 marks in CBSE Class 11 boards and roughly 1–2 questions in JEE Main every year — making it a high-reward chapter if you build it right.

The key insight most students miss: all four conics are described by a single general second-degree equation. Once you understand the structure, you stop memorising four separate chapters and start seeing one unified idea.

Key Terms and Definitions

Focus — A special fixed point inside the conic. Every point on the conic has a defined distance relationship to the focus (or foci).

Directrix — A fixed line outside the conic. The ratio of distance from any point on the conic to the focus versus its distance to the directrix is the eccentricity.

Eccentricity (e) — The defining number of a conic:

$e = 0$ : Circle
$0 < e < 1$ : Ellipse
$e = 1$ : Parabola
$e > 1$ : Hyperbola

Vertex — The point(s) where the conic meets its axis of symmetry.

Latus Rectum — The chord passing through the focus, perpendicular to the major axis. Its length appears repeatedly in PYQs — know this formula cold.

Centre — For circle, ellipse, and hyperbola: the midpoint of the line joining the two foci. Parabola has no centre (it’s not bounded symmetrically on both sides).

The Four Conics — Concepts and Formulas

Circle

x^2 + y^2 = r^2 \quad \text{(centre at origin)}

(x-h)^2 + (y-k)^2 = r^2 \quad \text{(centre at } (h,k)\text{)}

A circle is the locus of all points equidistant from the centre. The distance is $r$ , the radius.

General form: $x^2 + y^2 + 2gx + 2fy + c = 0$ , where centre $= (-g, -f)$ and $r = \sqrt{g^2 + f^2 - c}$ .

Students write the centre as $(g, f)$ instead of $(-g, -f)$ from the general form. The signs flip — half the class drops marks here every year.

Parabola

y^2 = 4ax \quad \text{(opens right, focus at } (a,0)\text{)}

y^2 = -4ax \quad \text{(opens left)}

x^2 = 4ay \quad \text{(opens up, focus at } (0,a)\text{)}

x^2 = -4ay \quad \text{(opens down)}

For $y^2 = 4ax$ :

Focus: $(a, 0)$
Directrix: $x = -a$
Vertex: $(0, 0)$
Axis: $x$ -axis
Latus rectum length: $4a$

The latus rectum length for parabola is always $4a$ — same as the coefficient in the standard form. Hard to forget once you see it this way.

Ellipse

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad a > b > 0

c^2 = a^2 - b^2, \quad e = \frac{c}{a} < 1

Key properties for $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ with $a > b$ :

Property	Value
Foci	$(\pm c, 0)$
Vertices	$(\pm a, 0)$
Co-vertices	$(0, \pm b)$
Length of major axis	$2a$
Length of minor axis	$2b$
Latus rectum length	$\dfrac{2b^2}{a}$
Directrices	$x = \pm \dfrac{a}{e}$

Vertical ellipse: When $b > a$ , the major axis is along the $y$ -axis. Swap the roles: foci at $(0, \pm c)$ with $c^2 = b^2 - a^2$ .

JEE Main frequently asks you to find the equation of an ellipse given eccentricity + one focus or given latus rectum + relationship between $a$ and $b$ . Always write down what’s given, identify $a$ , $b$ , $c$ , $e$ explicitly before forming the equation.

Hyperbola

\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1

c^2 = a^2 + b^2, \quad e = \frac{c}{a} > 1

Key properties:

Property	Value
Foci	$(\pm c, 0)$
Vertices	$(\pm a, 0)$
Latus rectum	$\dfrac{2b^2}{a}$
Asymptotes	$y = \pm \dfrac{b}{a}x$
Directrices	$x = \pm \dfrac{a}{e}$

Conjugate hyperbola: $\dfrac{y^2}{b^2} - \dfrac{x^2}{a^2} = 1$ — foci on $y$ -axis.

Rectangular hyperbola: $xy = c^2$ (asymptotes are the coordinate axes). This form appears in JEE Advanced more than students expect.

For hyperbola, $c^2 = a^2 + b^2$ (addition), but for ellipse $c^2 = a^2 - b^2$ (subtraction). Writing it wrong on exam day is extremely common — write this distinction at the top of your conic sections notes page.

Solved Examples

Example 1 — Easy (CBSE Level)

Find the equation of a parabola with vertex at the origin, axis along the positive $x$ -axis, and passing through $(3, 6)$ .

Standard form: $y^2 = 4ax$ .

Substitute $(3, 6)$ : $36 = 4a(3) \Rightarrow 36 = 12a \Rightarrow a = 3$ .

Answer: $y^2 = 12x$

Example 2 — Medium (JEE Main Level)

Find the eccentricity of the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$ . Also find the length of the latus rectum.

Here $a^2 = 25$ , $b^2 = 9$ , so $a = 5$ , $b = 3$ .

c^2 = a^2 - b^2 = 25 - 9 = 16 \Rightarrow c = 4

e = \frac{c}{a} = \frac{4}{5}

Latus rectum $= \dfrac{2b^2}{a} = \dfrac{2 \times 9}{5} = \dfrac{18}{5}$

Example 3 — Medium (JEE Main Level)

The foci of a hyperbola are $(\pm 5, 0)$ and the length of its latus rectum is $\dfrac{16}{3}$ . Find the equation.

Foci at $(\pm 5, 0)$ means $c = 5$ , so $c^2 = 25$ .

Latus rectum $= \dfrac{2b^2}{a} = \dfrac{16}{3}$ , so $b^2 = \dfrac{8a}{3}$ .

Since $c^2 = a^2 + b^2$ : $25 = a^2 + \dfrac{8a}{3}$

$75 = 3a^2 + 8a \Rightarrow 3a^2 + 8a - 75 = 0$

$(3a + 25)(a - 3) = 0 \Rightarrow a = 3$ (taking positive value)

$b^2 = 25 - 9 = 16$

Answer: $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$

Example 4 — Hard (JEE Advanced Level)

A variable point $P(x, y)$ moves such that its distance from $(2, 0)$ is always $\dfrac{3}{4}$ times its distance from the line $x = \dfrac{8}{3}$ . Find the locus and identify the conic.

Distance from focus $(2, 0)$ = $\sqrt{(x-2)^2 + y^2}$

Distance from directrix $x = \dfrac{8}{3}$ = $\left|x - \dfrac{8}{3}\right|$

Given ratio (eccentricity): $e = \dfrac{3}{4} < 1$ , so this is an ellipse.

\sqrt{(x-2)^2 + y^2} = \frac{3}{4}\left(x - \frac{8}{3}\right)

Squaring: $(x-2)^2 + y^2 = \dfrac{9}{16}\left(x - \dfrac{8}{3}\right)^2$

$(x-2)^2 + y^2 = \dfrac{9}{16}\left(\dfrac{3x-8}{3}\right)^2 = \dfrac{(3x-8)^2}{16}$

$16(x-2)^2 + 16y^2 = (3x-8)^2$

$16(x^2 - 4x + 4) + 16y^2 = 9x^2 - 48x + 64$

$16x^2 - 64x + 64 + 16y^2 = 9x^2 - 48x + 64$

$7x^2 + 16y^2 - 16x = 0$

$7(x^2 - \tfrac{16}{7}x) + 16y^2 = 0$

$7\left(x - \tfrac{8}{7}\right)^2 + 16y^2 = \tfrac{64}{7}$

\frac{\left(x - \frac{8}{7}\right)^2}{\frac{64}{49}} + \frac{y^2}{\frac{4}{7}} = 1

This is an ellipse with centre $\left(\dfrac{8}{7}, 0\right)$ .

This locus-based approach — starting from the eccentricity definition — is the standard JEE Advanced method. Practice this template: identify focus, identify directrix, write the ratio, square and simplify. Every locus of conic problem follows the same skeleton.

Exam-Specific Tips

CBSE Class 11 Boards

The CBSE paper typically has one 3-mark question (find equation given conditions) and one 4-mark question (latus rectum, eccentricity, foci). Standard forms are all you need — no locus derivations.

Marking scheme insight: you get 1 mark for the correct standard form, 1 mark for substituting the condition, and 1-2 marks for the final answer. Show your work — a wrong final answer with correct method gets partial credit.

JEE Main

Weightage: 1-2 questions per paper, roughly 4 marks. Parabola and ellipse appear more often than hyperbola. JEE Main 2024 Session 1 had a question on the normal to a parabola — a topic just outside Class 11 scope, but worth being aware of if you’re targeting 90+ percentile.

Focus areas: finding equations from given conditions (eccentricity + vertex/focus), length of latus rectum, intersection of line with conic (discriminant condition for tangency).

JEE Main 2023 had a straightforward ellipse question: given $e = \frac{1}{2}$ and one focus at $(1, 0)$ , find $a$ and $b$ . Solved in under 90 seconds if you know $c = ae$ cold. Drill the relationships $c = ae$ and $c^2 = a^2 - b^2$ until they’re automatic.

JEE Advanced

Expect parametric forms, normals, tangents, chord of contact, and properties of the rectangular hyperbola. The level of difficulty is significantly higher, but the foundation is the same — strong standard forms and eccentricity intuition.

Common Mistakes to Avoid

Mistake 1 — Swapping $a$ and $b$ in ellipse. Always check: $a > b$ for the horizontal major axis. If $b > a$ in your answer, double-check whether foci should be on the $y$ -axis instead.

Mistake 2 — Using $c^2 = a^2 - b^2$ for hyperbola. This is the ellipse formula. For hyperbola: $c^2 = a^2 + b^2$ . Mixing these two costs marks in JEE every year.

Mistake 3 — Forgetting the $\pm$ for foci. Foci always come in pairs (except parabola, which has one). Writing just $(c, 0)$ instead of $(\pm c, 0)$ loses a mark in CBSE.

Mistake 4 — Latus rectum confusion. The formula $\dfrac{2b^2}{a}$ gives the length of the latus rectum for ellipse and hyperbola. For parabola $y^2 = 4ax$ , it’s simply $4a$ . Many students apply the $\dfrac{2b^2}{a}$ formula to parabolas — it doesn’t apply.

Mistake 5 — General form to standard form errors. When converting $x^2 + y^2 + 2gx + 2fy + c = 0$ to standard form by completing the square, students forget to move $c$ to the right side before completing. Always: rearrange first, then complete the square.

Practice Questions

Q1. Find the equation of a circle with centre $(3, -4)$ passing through $(1, 2)$ .

Distance from $(3,-4)$ to $(1,2)$ : $r = \sqrt{(3-1)^2 + (-4-2)^2} = \sqrt{4+36} = \sqrt{40}$

Equation: $(x-3)^2 + (y+4)^2 = 40$

Q2. Find the focus, directrix, and latus rectum of $x^2 = -16y$ .

This is $x^2 = -4ay$ form, so $4a = 16 \Rightarrow a = 4$ .

Opens downward: Focus $= (0, -4)$ , Directrix: $y = 4$ , Latus rectum $= 4a = 16$ .

Q3. For the ellipse $\dfrac{x^2}{16} + \dfrac{y^2}{7} = 1$ , find the eccentricity, foci, and length of latus rectum.

$a^2 = 16$ , $b^2 = 7$ , $c^2 = 16 - 7 = 9$ , $c = 3$ .

$e = \dfrac{3}{4}$ , Foci: $(\pm 3, 0)$ , Latus rectum $= \dfrac{2 \times 7}{4} = \dfrac{7}{2}$

Q4. Find the equation of the ellipse with foci $(\pm 3, 0)$ and passing through $(4, 1)$ .

$c = 3$ , so $c^2 = 9$ . Standard form: $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ with $b^2 = a^2 - 9$ .

Substitute $(4, 1)$ : $\dfrac{16}{a^2} + \dfrac{1}{a^2 - 9} = 1$

$16(a^2-9) + a^2 = a^2(a^2-9)$

$17a^2 - 144 = a^4 - 9a^2$

$a^4 - 26a^2 + 144 = 0$

$(a^2 - 18)(a^2 - 8) = 0$

Since $a^2 > c^2 = 9$ : $a^2 = 18$ , $b^2 = 9$ .

Answer: $\dfrac{x^2}{18} + \dfrac{y^2}{9} = 1$

Q5. Find the equation of the hyperbola with vertices $(\pm 3, 0)$ and eccentricity $\dfrac{5}{3}$ .

$a = 3$ , $e = \dfrac{5}{3}$ , so $c = ae = 5$ .

$b^2 = c^2 - a^2 = 25 - 9 = 16$

Answer: $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$

Q6. A circle passes through $(0, 0)$ , $(4, 0)$ , and $(0, 3)$ . Find its equation.

General form: $x^2 + y^2 + 2gx + 2fy + c = 0$

Through $(0,0)$ : $c = 0$

Through $(4,0)$ : $16 + 8g = 0 \Rightarrow g = -2$

Through $(0,3)$ : $9 + 6f = 0 \Rightarrow f = -\dfrac{3}{2}$

Answer: $x^2 + y^2 - 4x - 3y = 0$

Q7. Find the length of the latus rectum and the equation of the directrices of the hyperbola $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$ .

$a^2 = 9$ , $b^2 = 16$ , $c^2 = 25$ , $c = 5$ , $e = \dfrac{5}{3}$ .

Latus rectum $= \dfrac{2b^2}{a} = \dfrac{32}{3}$

Directrices: $x = \pm \dfrac{a}{e} = \pm \dfrac{9}{5}$

Q8. Show that the locus of a point which moves such that the sum of its distances from $(-1, 0)$ and $(1, 0)$ is always $4$ is an ellipse. Find its equation.

Sum of distances from two foci $= 4 = 2a \Rightarrow a = 2$ .

Foci at $(\pm 1, 0) \Rightarrow c = 1$ .

$b^2 = a^2 - c^2 = 4 - 1 = 3$

Answer: $\dfrac{x^2}{4} + \dfrac{y^2}{3} = 1$

This is an ellipse because the sum of focal distances is constant, which is the defining property of an ellipse.

FAQs

What is the difference between a parabola and a hyperbola?

A parabola has eccentricity exactly $1$ — one focus, one directrix, one branch. A hyperbola has $e > 1$ — two foci, two directrices, two branches that open in opposite directions. The key algebraic difference: parabola has only one squared term ( $y^2$ or $x^2$ ), while hyperbola has two ( $x^2$ and $y^2$ ) with opposite signs.

Why does $c^2 = a^2 + b^2$ for hyperbola but $c^2 = a^2 - b^2$ for ellipse?

In an ellipse, the foci lie inside the ellipse, between the centre and the vertices — so $c < a$ , meaning $c^2 < a^2$ . In a hyperbola, the foci lie outside the vertices, so $c > a$ . Geometrically, $b$ is the semi-axis perpendicular to the transverse axis, and it “adds” to $a$ in the hyperbola case instead of being subtracted.

How do I identify a conic from the general second-degree equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ ?

For the NCERT/CBSE syllabus (no $xy$ term, $B = 0$ ): if $A = C$ , it’s a circle; if one of $A$ or $C$ is zero, it’s a parabola; if $A$ and $C$ have the same sign (but $A \neq C$ ), it’s an ellipse; if $A$ and $C$ have opposite signs, it’s a hyperbola.

What is the latus rectum and why does it keep appearing in exam questions?

The latus rectum is the chord through the focus perpendicular to the major axis. It’s a natural “size descriptor” of the conic — you can fully determine an ellipse or hyperbola if you know $a$ and the latus rectum length $\dfrac{2b^2}{a}$ . Examiners love it because it links $a$ , $b$ , and $c$ in one expression.

Can the same equation represent different conics?

Yes — the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$ can represent a real circle, a point circle ( $r = 0$ ), or an imaginary circle depending on the values of $g$ , $f$ , $c$ . Real circle: $g^2 + f^2 - c > 0$ . Point circle: $g^2 + f^2 - c = 0$ . Imaginary: $g^2 + f^2 - c < 0$ .

How many questions from conic sections appear in JEE Main?

Typically 1–2 questions per paper (4–8 marks total). Over the last three years, parabola and ellipse have appeared more frequently than hyperbola. The chapter has roughly 3–4% weightage in JEE Main — not the highest, but very predictable in the type of questions asked.

Is the rectangular hyperbola $xy = c^2$ in the Class 11 syllabus?

It’s mentioned briefly in NCERT but not emphasised for boards. For JEE Advanced, it becomes important — especially properties like the normal and parametric form $(ct, c/t)$ . For JEE Main, focus on the standard forms first; rectangular hyperbola is a bonus.