Conic Sections ISC Class 11: Parabola, Ellipse and Hyperbola

What Conic Sections is about

Weightage: Conic Sections belongs to the unit Coordinate Geometry, which CISCE weights at 20 of the 80 marks in the ISC Class XI Mathematics Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Mathematics syllabus, examination year 2028, Class XI unit weightage

Key concepts

Sections of a cone and degenerate conics

A double right circular cone has a vertex V, a vertical axis and semi-vertical angle α. Let a cutting plane make angle β with the axis. If the plane misses V: β = 90° gives a circle; α < β < 90° an ellipse; β = α (plane parallel to a generator) a parabola; 0 ≤ β < α a hyperbola (the plane cuts both nappes, giving two branches). If the plane passes through V, the section degenerates: α < β ≤ 90° gives a point; β = α a straight line (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 with the axis of (i) 60° (ii) 30° (iii) 20°. What if the plane in (iii) passes through the vertex?

Try it first, then show the solutionSolution

1. (i) 30° < 60° < 90°: ellipse
2. (ii) β = α: parabola
3. (iii) β < α: hyperbola; through the vertex: a pair of intersecting lines

Answer: Ellipse; parabola; hyperbola (pair of lines if through the vertex)

Focus-directrix definition PS = e·PL

Given a focus S, a directrix and an eccentricity e, take P(x, y) on the conic, write PS² = e² · PL² with PL found by the point-to-line distance formula, and simplify. The axis is the line through S perpendicular to the directrix; a vertex is where the conic cuts the axis. The latus rectum is the focal chord perpendicular to the axis. This one definition generates all three curves.

Worked example

A point moves so that its distance from (3, 0) is half its distance from the line x = 12. Find its path.

Try it first, then show the solutionSolution

1. (x − 3)² + y² = ¼(x − 12)²
2. 4x² − 24x + 36 + 4y² = x² − 24x + 144 ⇒ 3x² + 4y² = 108
3. x²/36 + y²/27 = 1, an ellipse with e = 1/2 (check: b² = 36(1 − 1/4) = 27 ✓)

Answer: x²/36 + y²/27 = 1

Parabola: four standard forms

With focus S(a, 0) and directrix x = −a (a > 0), PS = PL gives y² = 4ax. Its vertex is (0, 0), axis y = 0, it opens towards the focus (lies in Q I and Q IV), and the latus rectum has length 4a with ends (a, ±2a). The other forms follow by symmetry: y² = −4ax (focus (−a, 0), directrix x = a, opens left, Q II and Q III); x² = 4ay (focus (0, a), directrix y = −a, opens up); x² = −4ay (focus (0, −a), directrix y = a, opens down). The squared variable tells you the axis is the other one.

Worked example

For y² = −12x, find the focus, directrix, axis and latus rectum, and say where it lies.

Try it first, then show the solutionSolution

1. 4a = 12 ⇒ a = 3, negative x: opens left
2. Focus (−3, 0); directrix x = 3; axis y = 0
3. Latus rectum 12, ends (−3, ±6); lies in Q II and Q III

Answer: Focus (−3, 0), directrix x = 3, LR 12

Finding a parabola from data

If the vertex is at the origin and the axis is a coordinate axis, choose the form by the direction of opening, then use a given point or the focus to find a. If only a focus and directrix are given (not necessarily parallel to an axis), use PS = PL directly. The vertex is always midway between the focus and the directrix along the axis.

Worked example

Find the parabola with vertex at the origin, axis along the y-axis, passing through (4, −2).

Try it first, then show the solutionSolution

1. The point has negative y, so it opens downwards: x² = −4ay
2. 16 = −4a(−2) ⇒ a = 2

Answer: x² = −8y; focus (0, −2)

Ellipse: standard form and features

For x²/a² + y²/b² = 1 with a > b: centre (0, 0); major axis along x of length 2a, minor axis along y of length 2b; vertices (±a, 0); foci (±ae, 0) with b² = a²(1 − e²) (so ae = √(a² − b²)); directrices x = ±a/e; latus rectum 2b²/a. If a < b, the major axis is along y: vertices (0, ±b), foci (0, ±be) with a² = b²(1 − e²), directrices y = ±b/e, latus rectum 2a²/b. Always identify the larger denominator first. A rough sketch is an oval through the four axis end-points with the foci inside on the major axis.

Worked example

For 9x² + 25y² = 225, find a, b, e, the foci, directrices and latus rectum.

Try it first, then show the solutionSolution

1. Divide by 225: x²/25 + y²/9 = 1; a = 5, b = 3 (major axis along x)
2. 9 = 25(1 − e²) ⇒ e = 4/5; foci (±4, 0)
3. Directrices x = ±a/e = ±25/4; LR = 2 × 9/5 = 18/5

Answer: e = 4/5; foci (±4, 0); x = ±25/4; LR 18/5

Ellipse: focal property and equations from data

For any point P on the ellipse, PS = e·PM and PS′ = e·PM′, where M, M′ are the feet of perpendiculars on the two directrices; since PM + PM′ = 2a/e, we get SP + S′P = 2a, the sum of focal distances equals the major axis. Hence the 'string and two pins' construction. To find an ellipse from data, use any two of: a, b, ae, e, a/e, latus rectum, a point on the curve, together with b² = a²(1 − e²).

Worked example

Find the ellipse with vertices (±10, 0) and foci (±6, 0), and verify SP + S′P = 2a at an end of the minor axis.

Try it first, then show the solutionSolution

1. a = 10, ae = 6 ⇒ b² = 100 − 36 = 64
2. Ellipse x²/100 + y²/64 = 1
3. At P(0, 8): SP = S′P = √(36 + 64) = 10, sum 20 = 2a ✓

Answer: x²/100 + y²/64 = 1

Hyperbola: standard form and features

For x²/a² − y²/b² = 1: centre (0, 0); transverse axis along x, length 2a; conjugate axis along y, length 2b; vertices (±a, 0); foci (±ae, 0) with b² = a²(e² − 1) (so ae = √(a² + b²)); directrices x = ±a/e; latus rectum 2b²/a. Here a and b need not be in any order. If the x² term is negative, y²/a² − x²/b² = 1 (in the usual convention), the transverse axis is along y: vertices (0, ±a), foci (0, ±ae), directrices y = ±a/e. A rough sketch has two separate branches opening away from the centre.

Worked example

For 16x² − 9y² = 144, find a, b, e, the foci, directrices and latus rectum.

Try it first, then show the solutionSolution

1. Divide by 144: x²/9 − y²/16 = 1; a = 3, b = 4
2. 16 = 9(e² − 1) ⇒ e = 5/3; foci (±5, 0)
3. Directrices x = ±9/5; LR = 2 × 16/3 = 32/3

Answer: e = 5/3; foci (±5, 0); x = ±9/5; LR 32/3

Hyperbola: focal property and equations from data

For a point P on a hyperbola, the difference of its distances from the foci is constant: |SP − S′P| = 2a, the length of the transverse axis. To find a hyperbola from data, decide which axis the foci lie on (that is the transverse axis), then use c = ae with c² = a² + b². Questions often give the foci with vertices, eccentricity, the conjugate axis or a point on the curve.

Worked example

Find the hyperbola with foci (0, ±10) and vertices (0, ±6), and check the focal property at a point with y = 10.

Try it first, then show the solutionSolution

1. Transverse axis along y: a = 6, ae = 10, b² = 100 − 36 = 64; y²/36 − x²/64 = 1
2. y = 10: x² = 64(100/36 − 1) = 4096/36, x = 32/3
3. Distances from (0, 10) and (0, −10): 32/3 and √(1024/9 + 400) = 68/3; difference 36/3 = 12 = 2a ✓

Answer: y²/36 − x²/64 = 1

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: Standard Forms
• Parabola: Standard Forms and Equations

Conic Sections formula sheet

Focus-directrix

$PS=e\cdot PL$

e = 1 parabola; e < 1 ellipse; e > 1 hyperbola.

Parabola y² = 4ax

$S(a,0),x=-a,\text{LR}=4a$

x² = 4ay: S(0, a), y = −a.

Ellipse relation

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

For a > b; ae = √(a² − b²).

Ellipse features (a > b)

$S(\pm ae,0),x=\pm \frac{a}{e},\text{LR}=\frac{2b^2}{a}$

Vertices (±a, 0).

Ellipse focal property

$SP+S^{\prime }P=2a$

Sum of focal distances = major axis.

Hyperbola relation

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

ae = √(a² + b²).

Hyperbola features

$S(\pm ae,0),x=\pm \frac{a}{e},\text{LR}=\frac{2b^2}{a}$

Transverse axis 2a, conjugate axis 2b.

Hyperbola focal property

$|SP-S^{\prime }P|=2a$

Difference of focal distances = transverse axis.

Conic Sections: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

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.

• e decides the curve

  e = 1 parabola (equal), e < 1 ellipse (closed), e > 1 hyperbola (opens out).

  Use it when: Classifying from PS = e·PL.

• Squared variable, other axis

  In y² = 4ax the y is squared and the axis is the x-axis; the sign of the other side gives the direction of opening.

  Use it when: Parabola orientation.

• Ellipse minus, hyperbola plus

  Ellipse: c² = a² − b²; hyperbola: c² = a² + b², where c = ae.

  Use it when: Finding foci.

• Directrix is a over e

  Foci at ±ae, directrices at ±a/e: e multiplies for foci, divides for directrices.

  Use it when: Not mixing ae and a/e.

• Sum for ellipse, difference for hyperbola

  SP + S′P = 2a; |SP − S′P| = 2a.

  Use it when: Focal properties.

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 D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 1 + 2 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 as the x-denominator when the major axis is along y

  Students compute e from the smaller denominator as if it were a².

  Wrong working

  x²/16 + y²/25 = 1: 25 = 16(1 − e²).

  Right working

  Larger denominator 25 is under y²: 16 = 25(1 − e²) ⇒ e = 3/5.

  ✓Do this: Identify the larger denominator before anything else.

  Usually costs: Every later answer.

• ✗Mistake: Using the ellipse relation for a hyperbola

  b² = a²(1 − e²) is applied to both.

  Wrong working

  x²/9 − y²/16 = 1: 16 = 9(1 − e²).

  Right working

  16 = 9(e² − 1) ⇒ e = 5/3.

  ✓Do this: Hyperbola e > 1, so e² − 1.

  Usually costs: 2-3 marks.

• ✗Mistake: Wrong sign for the focus of x² = −4ay

  Students copy (0, a).

  Wrong working

  x² = −20y: focus (0, 5).

  Right working

  Opens downwards: focus (0, −5), directrix y = 5.

  ✓Do this: The focus is inside the curve, on the side it opens towards.

  Usually costs: 1 mark.

• ✗Mistake: Confusing ae with a/e

  Foci and directrices are both on the major axis.

  Wrong working

  Ellipse x²/25 + y²/9 = 1: directrices x = ±4.

  Right working

  Foci x = ±ae = ±4; directrices x = ±a/e = ±25/4.

  ✓Do this: For an ellipse the directrix is further out than the vertex.

  Usually costs: 1 mark.

• ✗Mistake: Wrong latus rectum formula when a < b

  Students use 2b²/a for an ellipse with major axis along y.

  Wrong working

  x²/16 + y²/25 = 1: LR = 2 × 25/4.

  Right working

  LR = 2 × (semi-minor)²/(semi-major) = 2 × 16/5 = 32/5.

  ✓Do this: LR = 2(minor semi-axis)²/(major semi-axis).

  Usually costs: 1 mark.

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 eccentricity of every parabola is:
   (a) 0
   (b) 1
   (c) between 0 and 1
   (d) greater than 1

   Hint

   PS = PL.

   Answer

   (b)

2. Q2 · 1 mark · MCQExam level

   The length of the latus rectum of y² = 20x is:
   (a) 5
   (b) 10
   (c) 20
   (d) 40

   Hint

   4a.

   Answer

   (c)

3. Q3 · 1 mark · Very short answerExam level

   Write the equation of the directrix of x² = 8y.

   Hint

   a = 2.

   Answer

   y = −2

4. Q4 · 1 mark · Very short answerExam level

   Find the eccentricity of x²/25 + y²/16 = 1.

   Hint

   16 = 25(1 − e²).

   Answer

   3/5

5. Q5 · 2 marks · Short answerExam level

   Find the lengths of the major and minor axes of 9x² + 16y² = 144.

   Hint

   x²/16 + y²/9 = 1.

   Answer

   Major 8, minor 6

6. Q6 · 2 marks · Short answerHarder exam level

   Find the parabola with vertex at the origin and focus (0, −3).

   Hint

   Opens downwards.

   Answer

   x² = −12y

7. Q7 · 2 marks · Short answerHarder exam level

   Find the ellipse with vertices (±9, 0) and foci (±6, 0).

   Hint

   b² = a² − (ae)².

   Answer

   b² = 81 − 36 = 45: x²/81 + y²/45 = 1

8. Q8 · 3 marks · Short answerHarder exam level

   Find the hyperbola with foci (0, ±17) and conjugate axis of length 30.

   Hint

   Transverse axis along y; b = 15.

   Answer

   a² = 289 − 225 = 64: y²/64 − x²/225 = 1

9. Q9 · 3 marks · Short answerHarder exam level

   Find the parabola with vertex at the origin, axis along the x-axis, passing through (2, 3), and its focus.

   Hint

   y² = 4ax.

   Answer

   9 = 8a ⇒ a = 9/8: y² = 9x/2 (2y² = 9x); focus (9/8, 0)

10. Q10 · 3 marks · Short answerHarder exam level

    An ellipse has centre at the origin, a focus at (2, 0) and the corresponding directrix x = 8. Find its eccentricity and equation.

    Hint

    ae = 2 and a/e = 8.

    Answer

    Multiplying: a² = 16; dividing: e² = 1/4, e = 1/2; b² = 16 × 3/4 = 12: x²/16 + y²/12 = 1

11. Q11 · 3 marks · Short answerHarder exam level

    For 25x² − 9y² = 225, find a, b, e, the foci and the latus rectum.

    Hint

    x²/9 − y²/25 = 1.

    Answer

    a = 3, b = 5; e = √(1 + 25/9) = √34/3; foci (±√34, 0); LR = 2 × 25/3 = 50/3

12. Q12 · 2 marks · Short answerHarder exam level

    Find the focus, directrix and latus rectum of 3y² = −8x.

    Hint

    y² = −(8/3)x.

    Answer

    4a = 8/3, a = 2/3: focus (−2/3, 0), directrix x = 2/3, LR 8/3

13. Q13 · 5 marks · Long answerHarder exam level

    (a) Find the ellipse (major axis along the x-axis, centre at the origin) whose latus rectum is 5 and eccentricity 2/3. [3]
    (b) Find its foci. [1]
    (c) Find its directrices. [1]

    Hint

    2b²/a = 5 and b² = a²(1 − 4/9).

    Answer

    (a) b² = 5a²/9, so 2(5a²/9)/a = 10a/9 = 5 ⇒ a = 9/2; a² = 81/4, b² = 45/4: 4x²/81 + 4y²/45 = 1. (b) ae = 3: (±3, 0). (c) x = ±a/e = ±27/4.

14. Q14 · 3 marks · Short answerHOTS / competency

    Find the points on y² = 8x whose distance from the focus is 6.

    Hint

    Distance from focus = distance from directrix = x + 2.

    Answer

    x + 2 = 6 ⇒ x = 4; y² = 32 ⇒ y = ±4√2. Points (4, ±4√2).

15. Q15 · 3 marks · Short answerHOTS / competency

    Find the hyperbola with foci (±5, 0) that passes through (4, 3).

    Hint

    a² + b² = 25.

    Answer

    16/a² − 9/(25 − a²) = 1 ⇒ a⁴ − 50a² + 400 = 0 ⇒ a² = 10 (a² = 40 > 25 is impossible); b² = 15: x²/10 − y²/15 = 1

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.Focus-directrix definition of a conic?Show answer

   PS = e·PL.

2. 2.Focus and directrix of y² = 4ax?Show answer

   (a, 0); x = −a.

3. 3.Latus rectum of y² = 4ax?Show answer

   4a.

4. 4.Ellipse relation between a, b, e?Show answer

   b² = a²(1 − e²).

5. 5.Hyperbola relation between a, b, e?Show answer

   b² = a²(e² − 1).

6. 6.Directrices of x²/a² ± y²/b² = 1?Show answer

   x = ±a/e.

7. 7.Focal property of an ellipse?Show answer

   SP + S′P = 2a.

8. 8.Section when the plane is parallel to a generator (not through the vertex)?Show answer

   A parabola.

Quick revision checklist

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

• ☐Can you name the section for a given plane angle?
• ☐Can you find a conic from a focus, a directrix and e?
• ☐Can you state the focus, directrix and latus rectum of x² = −4ay?
• ☐Can you find e, foci and directrices of an ellipse with a < b?
• ☐Can you verify SP + S′P = 2a at a point?
• ☐Can you find e, foci and latus rectum of a hyperbola?
• ☐Can you find a hyperbola from its foci and a point?
• ☐Can you sketch all three curves roughly with features marked?

Conic Sections: frequently asked questions