Conic Sections Class 11: Circle, Parabola, Ellipse, Hyperbola and Solved Questions

What Conic Sections is about

Weightage: This chapter is part of Unit III, Coordinate Geometry, which the syllabus gives 12 of the 80 theory marks in the school annual exam (CBSE adds that there is no fixed chapter-wise weightage).Source: CBSE Mathematics (041) syllabus 2026-27, Class XI course structure, Unit III Coordinate Geometry

Key concepts

Sections of a cone and degenerate conics

NCERT section 10.2

Let a fixed vertical line l (the axis) and a line m through a point V on it make an angle α. Rotating m about l gives a double-napped cone with vertex V and semi-vertical angle α. Now cut it with a plane making an angle β with the axis.

Plane not through the vertex:

• β = 90°: circle
• α < β < 90°: ellipse
• β = α: parabola (plane parallel to a generator)
• 0 ≤ β < α: hyperbola (the plane cuts both nappes)

Plane through the vertex (degenerate conics):

• α < β ≤ 90°: a point
• β = α: a straight line (the plane touches the cone along a generator)
• 0 ≤ β < α: a pair of intersecting lines

Worked example

A cone has semi-vertical angle 30°. Name the section when a plane not through the vertex makes an angle of (a) 90° (b) 50° (c) 30° (d) 20° with the axis. What if the plane at 20° passes through the vertex? [2 marks]

Try it first, then show the solutionSolution

1. (a) circle (b) 30° < 50° < 90°: ellipse
2. (c) β = α: parabola (d) β < α: hyperbola; through the vertex at 20°: a pair of intersecting lines

Answer: Circle, ellipse, parabola, hyperbola; pair of lines.

Circle

NCERT section 10.3

A circle is the set of points in a plane at a fixed distance r (radius) from a fixed point (h, k) (centre). By the distance formula:
$(x-h{)}^2+(y-k{)}^2=r^2$
Centre at the origin: $x^2+y^2=r^2$.

Expanded form: $x^2+y^2+2gx+2fy+c=0$ has centre $(-g,-f)$ and radius $\sqrt{g^2+f^2-c}$ (real when $g^2+f^2-c>0$). To read it, complete the squares; if the coefficients of $x^2$ and $y^2$ are equal but not 1, divide first.

Point position: P is inside, on or outside the circle according as its distance from the centre is less than, equal to or greater than r.

Circle through given points with centre on a line: let the centre be (h, k) on the line, set equal distances to the points, and solve.

Worked example

Find the centre and radius of $x^2+y^2+8x-6y-24=0$. [2 marks]

Try it first, then show the solutionSolution

1. $(x^2+8x+16)+(y^2-6y+9)=24+16+9$
2. $(x+4{)}^2+(y-3{)}^2=49$: centre (−4, 3), radius 7

Answer: Centre (−4, 3), radius 7

Parabola

NCERT section 10.4

A parabola is the set of points in a plane equidistant from a fixed line (the directrix) and a fixed point not on it (the focus). The line through the focus perpendicular to the directrix is the axis; the point where the parabola meets its axis is the vertex (midway between focus and directrix).

Four standard forms (vertex at origin, a > 0):

Equation	Focus	Directrix	Opens
$y^2=4ax$	(a, 0)	x = −a	right
$y^2=-4ax$	(−a, 0)	x = a	left
$x^2=4ay$	(0, a)	y = −a	up
$x^2=-4ay$	(0, −a)	y = a	down

Latus rectum: the chord through the focus perpendicular to the axis; its length is 4a, with ends (a, ±2a) for $y^2=4ax$.

Reading the form: the squared variable tells you the axis of symmetry is the *other* axis; the sign of the non-squared side gives the direction of opening.

Worked example

For $x^2=-12y$, find the focus, axis, directrix and length of the latus rectum. [2 marks]

Try it first, then show the solutionSolution

1. Compare with $x^2=-4ay$: 4a = 12, a = 3; the parabola opens downward and is symmetric about the y-axis
2. Focus (0, −3); directrix y = 3; latus rectum 4a = 12

Answer: Focus (0, −3); axis: y-axis; directrix y = 3; latus rectum 12

Ellipse: definition and standard equation

NCERT section 10.5 and 10.5.3

An ellipse is the set of points P in a plane with $P{F}_1+P{F}_2=2a$ (constant), where $F_1$, $F_2$ are fixed points (the foci) and 2a is greater than the distance 2c between them.

• Centre: mid-point of the foci. Major axis: through the foci, length 2a; its ends are the vertices. Minor axis: perpendicular to it through the centre, length 2b.
• Relation: $a^2=b^2+c^2$ (from a point at the end of the minor axis, each focus is at distance a).

Standard forms (centre at origin, a > b):

• Foci on the x-axis: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, foci (±c, 0), vertices (±a, 0).
• Foci on the y-axis: $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$, foci (0, ±c), vertices (0, ±a).

Which axis is major? The one under the larger denominator.

The ellipse is symmetric about both axes, and every point satisfies $-a\le x\le a$, $-b\le y\le b$ (for the first form).

Worked example

Find the foci, vertices, lengths of the axes for $\frac{x^2}{25}+\frac{y^2}{9}=1$, and check the definition at the point $(3,\frac{12}{5})$. [3 marks]

Try it first, then show the solutionSolution

1. a = 5, b = 3 (25 > 9, major axis on x-axis); $c=\sqrt{25-9}=4$
2. Foci (±4, 0); vertices (±5, 0); major axis 10, minor axis 6
3. $P{F}_1+P{F}_2=\sqrt{1+5.76}+\sqrt{49+5.76}=2.6+7.4=10=2a$ ✓

Answer: Foci (±4, 0), vertices (±5, 0), axes 10 and 6.

Ellipse: eccentricity and latus rectum

NCERT section 10.5.2 and 10.5.4

Eccentricity: $e=\frac{c}{a}$, the ratio of the distance of a focus from the centre to the semi-major axis. For an ellipse $0<e<1$ (when c = 0 the foci coincide and the ellipse becomes a circle).

• e close to 0: nearly circular. e close to 1: long and thin.

Latus rectum: a chord through a focus perpendicular to the major axis. Its length is
$\frac{2b^2}{a}$
For $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the ends are $\left(\pm c,\pm \frac{b^2}{a}\right)$.

Building an equation from data: identify whether the foci/vertices lie on the x- or y-axis, find two of a, b, c, and get the third from $a^2=b^2+c^2$.

Worked example

For $9x^2+4y^2=36$, find the foci, vertices, eccentricity and latus rectum. [3 marks]

Try it first, then show the solutionSolution

1. Divide by 36: $\frac{x^2}{4}+\frac{y^2}{9}=1$; 9 > 4, so the major axis is the y-axis: a = 3, b = 2
2. $c=\sqrt{9-4}=\sqrt{5}$: foci $(0,\pm \sqrt{5})$, vertices (0, ±3)
3. $e=\frac{\sqrt{5}}{3}$; latus rectum $=\frac{2\times 4}{3}=\frac{8}{3}$

Answer: Foci $(0,\pm \sqrt{5})$; vertices (0, ±3); $e=\frac{\sqrt{5}}{3}$; latus rectum $\frac{8}{3}$

Hyperbola

NCERT section 10.6

A hyperbola is the set of points P in a plane with $|P{F}_1-P{F}_2|=2a$ (constant), where the foci $F_1$, $F_2$ are 2c apart and $2a<2c$.

• Transverse axis: through the foci; its ends are the vertices, 2a apart. Conjugate axis: perpendicular through the centre, length 2b.
• Relation: $c^2=a^2+b^2$ (note the plus sign).
• Eccentricity $e=\frac{c}{a}>1$. Latus rectum $=\frac{2b^2}{a}$.

Standard forms:

• Transverse axis on x-axis: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, foci (±c, 0), vertices (±a, 0).
• Transverse axis on y-axis: $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$, foci (0, ±c), vertices (0, ±a).

Which axis? The one whose term is positive, regardless of which denominator is larger. In a hyperbola a may be smaller than b.

If a = b the hyperbola is equilateral (rectangular), with $e=\sqrt{2}$.

Worked example

For $\frac{x^2}{9}-\frac{y^2}{16}=1$, find the foci, vertices, eccentricity and latus rectum. [3 marks]

Try it first, then show the solutionSolution

1. x-term positive: transverse axis on the x-axis; a = 3, b = 4
2. $c=\sqrt{9+16}=5$: foci (±5, 0), vertices (±3, 0)
3. $e=\frac{5}{3}$; latus rectum $=\frac{2\times 16}{3}=\frac{32}{3}$

Answer: Foci (±5, 0); vertices (±3, 0); $e=\frac{5}{3}$; latus rectum $\frac{32}{3}$

Maths in real life: Conic Sections

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Ellipse and Hyperbola: Foci, Eccentricity and Equations
• Parabola: Focus, Directrix and Latus Rectum

Conic Sections formula sheet

Circle

$(x-h{)}^2+(y-k{)}^2=r^2;x^2+y^2+2gx+2fy+c=0:\text{centre }(-g,-f),r=\sqrt{g^2+f^2-c}$

Parabola

$y^2=4ax:\text{focus }(a,0),\text{directrix }x=-a,\text{latus rectum }4a$

Ellipse

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b),c^2=a^2-b^2,e=\frac{c}{a}<1$

Hyperbola

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,c^2=a^2+b^2,e=\frac{c}{a}>1$

Latus rectum of ellipse and hyperbola

$\ell =\frac{2b^2}{a}$

Focal property

$\text{ellipse: }P{F}_1+P{F}_2=2a,\text{hyperbola: }|P{F}_1-P{F}_2|=2a$

How the school exam asks Conic Sections

The question patterns to expect, and the method that works for each.

Tricks to remember Conic Sections

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Ellipse minus, hyperbola plus

  Ellipse: $c^2=a^2-b^2$ (foci inside). Hyperbola: $c^2=a^2+b^2$ (foci outside the vertices).

  Use it when: Finding c.

• Bigger below, major axis; positive term, transverse axis

  Ellipse: larger denominator decides. Hyperbola: the positive term decides.

  Use it when: Orientation.

• 4a for parabola, 2b²/a for the others

  Latus rectum: parabola 4a; ellipse and hyperbola $\frac{2b^2}{a}$.

  Use it when: Latus rectum.

• e tells the shape

  e = 0 circle, 0 < e < 1 ellipse, e = 1 parabola, e > 1 hyperbola.

  Use it when: Classifying conics.

• Square the other one

  In $y^2=4ax$ the axis is the x-axis (the variable NOT squared).

  Use it when: Parabola orientation.

Conic Sections: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Conic Sections

The slips examiners see most often, and what to write instead.

• ✗Mistake: Taking a from the first denominator in an ellipse

  Assuming the x-term always has a².

  Wrong working

  $\frac{x^2}{4}+\frac{y^2}{9}=1$: a = 2, foci on x-axis

  Right working

  a = 3 (larger), foci $(0,\pm \sqrt{5})$

  ✓Do this: Compare the denominators first.

  Usually costs: Most of the question.

• ✗Mistake: Using $c^2=a^2-b^2$ for a hyperbola

  Mixing the two relations.

  ✓Do this: Hyperbola: $c^2=a^2+b^2$.

  Usually costs: 2 marks.

• ✗Mistake: Not dividing to get 1 on the right

  Reading a and b from $9x^2+4y^2=36$ directly.

  ✓Do this: Divide by the constant first to reach standard form.

  Usually costs: 1-2 marks.

• ✗Mistake: Wrong focus sign for $y^2=-4ax$ or $x^2=-4ay$

  Ignoring the minus.

  ✓Do this: Negative side → parabola opens towards the negative axis; focus on that side.

  Usually costs: 1 mark.

• ✗Mistake: Completing the square with unequal $x^2$, $y^2$ coefficients

  Forgetting to divide by the common coefficient.

  Wrong working

  $2x^2+2y^2-6x+8y=0$: centre (3, −4)

  Right working

  Divide by 2: $x^2+y^2-3x+4y=0$: centre $(\frac{3}{2},-2)$, radius $\frac{5}{2}$

  ✓Do this: Make the coefficients of $x^2$ and $y^2$ equal to 1 first.

  Usually costs: 1-2 marks.

• ✗Mistake: Giving e > 1 for an ellipse

  Dividing a by c instead of c by a.

  ✓Do this: $e=\frac{c}{a}$; check it is below 1 for an ellipse and above 1 for a hyperbola.

  Usually costs: 1 mark.

NCERT exercise guide: Conic Sections

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Practice questions on Conic Sections

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The centre and radius of $2x^2+2y^2-6x+8y=0$ are:
   (a) (3, −4), 5
   (b) $(\frac{3}{2},-2)$, $\frac{5}{2}$
   (c) (−3, 4), 5
   (d) $(-\frac{3}{2},2)$, $\frac{5}{2}$

   Hint

   Divide by 2.

   Answer

   (b)

2. Q2 · 1 mark · MCQExam level

   The length of the latus rectum of $\frac{x^2}{64}+\frac{y^2}{36}=1$ is:
   (a) 9
   (b) $\frac{9}{2}$
   (c) 12
   (d) 18

   Hint

   $\frac{2b^2}{a}$.

   Answer

   (a): $\frac{2\times 36}{8}=9$

3. Q3 · 1 mark · Assertion-reasonExam level

   Assertion (A): The eccentricity of every hyperbola is greater than 1.
   Reason (R): For a hyperbola, $c^2=a^2+b^2$, so c > a.
   (a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

   Hint

   e = c/a.

   Answer

   (a)

4. Q4 · 1 mark · Very short answerExam level

   Does the point (4, −3.5) lie inside, on or outside the circle $x^2+y^2=25$?

   Hint

   Compare with 25.

   Answer

   $16+12.25=28.25>25$: outside.

5. Q5 · 2 marks · Very short answerHarder exam level

   Find the equation of the parabola with focus (0, −4) and directrix y = 4.

   Hint

   Opens downward.

   Answer

   a = 4: $x^2=-16y$.

6. Q6 · 2 marks · Very short answerHarder exam level

   Find the ellipse with foci (0, ±4) and semi-major axis a = 5.

   Hint

   $b^2=a^2-c^2$.

   Answer

   $b^2=25-16=9$; major axis on the y-axis: $\frac{x^2}{9}+\frac{y^2}{25}=1$.

7. Q7 · 2 marks · Very short answerHarder exam level

   Find the hyperbola with vertices (±3, 0) and foci (±4, 0).

   Hint

   $b^2=c^2-a^2$.

   Answer

   $b^2=16-9=7$: $\frac{x^2}{9}-\frac{y^2}{7}=1$.

8. Q8 · 3 marks · Short answerHarder exam level

   Find the foci, vertices, eccentricity and latus rectum of $25x^2+4y^2=100$.

   Hint

   Divide by 100.

   Answer

   $\frac{x^2}{4}+\frac{y^2}{25}=1$: a = 5 (y-axis), b = 2, $c=\sqrt{21}$. Foci $(0,\pm \sqrt{21})$, vertices (0, ±5), $e=\frac{\sqrt{21}}{5}$, latus rectum $\frac{2\times 4}{5}=\frac{8}{5}$.

9. Q9 · 3 marks · Short answerHarder exam level

   Find the equation of the circle passing through (0, 0) and cutting intercepts 4 and 6 on the positive x- and y-axes.

   Hint

   It passes through (4, 0) and (0, 6).

   Answer

   $x^2+y^2+2gx+2fy=0$ (through origin); (4, 0): $16+8g=0$, g = −2; (0, 6): $36+12f=0$, f = −3. So $x^2+y^2-4x-6y=0$.

10. Q10 · 3 marks · Short answerHOTS / competency

    A point moves so that the difference of its distances from (0, 5) and (0, −5) is always 8. Find the equation of its path.

    Hint

    Hyperbola with foci on the y-axis.

    Answer

    2a = 8, a = 4, c = 5, $b^2=9$: $\frac{y^2}{16}-\frac{x^2}{9}=1$.

11. Q11 · 2 marks · Very short answerHOTS / competency

    Show that an ellipse with eccentricity e and major axis 2a has latus rectum $2a(1-e^2)$, and check with $\frac{x^2}{25}+\frac{y^2}{9}=1$.

    Hint

    $b^2=a^2-c^2=a^2(1-e^2)$.

    Answer

    $\frac{2b^2}{a}=\frac{2a^2(1-e^2)}{a}=2a(1-e^2)$. Check: a = 5, e = 0.8: $10(1-0.64)=3.6=\frac{18}{5}$ ✓

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.What section does a plane parallel to a generator give?Show answer

   A parabola.

2. 2.Centre and radius of $x^2+y^2-4x+6y-3=0$?Show answer

   (2, −3) and 4.

3. 3.Focus of $y^2=20x$?Show answer

   (5, 0).

4. 4.Latus rectum of $y^2=4ax$?Show answer

   4a.

5. 5.Relation between a, b, c for an ellipse?Show answer

   $a^2=b^2+c^2$.

6. 6.Eccentricity range for a hyperbola?Show answer

   e > 1.

7. 7.Foci of $\frac{x^2}{9}-\frac{y^2}{16}=1$?Show answer

   (±5, 0).

8. 8.What is the eccentricity of a circle?Show answer

   0.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you name the conic for each angle of the cutting plane?
• ☐Can you write a circle from centre and radius and read centre and radius back?
• ☐Can you find a circle through two points with centre on a line?
• ☐Can you state focus, directrix and latus rectum for all four parabolas?
• ☐Can you find a parabola through a point with a given axis?
• ☐Can you find foci, vertices, e and latus rectum of any standard ellipse?
• ☐Can you find an ellipse from vertices and foci, or from two points?
• ☐Can you do the same for a hyperbola?
• ☐Can you explain why $e<1$ for an ellipse and $e>1$ for a hyperbola?
• ☐Can you solve an arch or reflector problem?

Conic Sections: frequently asked questions