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Maths · Class 11 · Chapter 10

Conic Sections

Circles, parabolas, ellipses and hyperbolas are all slices of a double cone. For exams, what you need is the standard equation of each and the handful of quantities attached to it: foci, vertices, eccentricity and latus rectum. A clean sketch prevents most sign errors.

In this chapter: sections of a cone, the circle in standard and general form, the parabola and its four standard forms, the ellipse, the hyperbola, and for each conic its focus, vertex, axes, eccentricity and latus rectum.

Sections of a cone

Take a double-napped right circular cone with a vertical axis and semi-vertical angle α. Cut it with a plane that does not pass through the vertex and makes an angle β with the axis.

Angle of the planeSection
β = 90°Circle
α < β < 90°Ellipse
β = α (plane parallel to a generator)Parabola
0° ≤ β < αHyperbola (the plane cuts both nappes)

If the plane passes through the vertex, we get degenerate conics: a point (α < β ≤ 90°), a single straight line (β = α), or a pair of intersecting lines (0 ≤ β < α).

Circle

A circle is the set of points in a plane at a fixed distance r (the radius) from a fixed point (h, k) (the centre).

(x − h)² + (y − k)² = r²x² + y² = r²centre at the originx² + y² + 2gx + 2fy + c = 0: centre (−g, −f), radius √(g² + f² − c)

The general form represents a real circle only when g² + f² − c > 0. Notice that in the equation of a circle the coefficients of x² and y² are equal and there is no xy term.

Worked example: Find the centre and radius of x² + y² + 8x + 10y − 8 = 0.
Solution: Complete the squares: (x² + 8x + 16) + (y² + 10y + 25) = 8 + 16 + 25, so (x + 4)² + (y + 5)² = 49. The centre is (−4, −5) and the radius is 7.

Parabola

A parabola is the set of all points in a plane that are equidistant from a fixed line (the directrix) and a fixed point (the focus) not on that line. The line through the focus perpendicular to the directrix is the axis; the point where the parabola meets its axis is the vertex. The latus rectum is the chord through the focus perpendicular to the axis.

Parabola y² = 4ax with focus and directrixwww.iitmedicoguide.comxydirectrixx = −alatus rectum = 4aP(x, y)MS(a, 0)Oy² = 4ax (a > 0)vertex: O(0, 0)focus: S(a, 0)directrix: x = −aaxis: y = 0PS = PM for everypoint P on the curve(eccentricity e = 1)www.iitmedicoguide.com
For y² = 4ax the focus is S(a, 0) and the directrix is x = −a. Every point P on the curve is as far from S as from the directrix, and the latus rectum through S has length 4a.
Equation (a > 0)OpensFocusDirectrixAxis
y² = 4axright(a, 0)x = −ay = 0
y² = −4axleft(−a, 0)x = ay = 0
x² = 4ayup(0, a)y = −ax = 0
x² = −4aydown(0, −a)y = ax = 0

In all four the vertex is the origin and the latus rectum has length 4a. A quick way to read the table: the squared variable tells you the axis of symmetry is the other axis, and the sign on the right tells you the direction of opening. For y² = 8x, 4a = 8 so a = 2: focus (2, 0), directrix x = −2, latus rectum 8.

Ellipse

An ellipse is the set of points in a plane the sum of whose distances from two fixed points (the foci) is a constant, taken as 2a. The midpoint of the foci is the centre.

x²/a² + y²/b² = 1, a > b > 0major axis along the x-axisc² = a² − b², foci (±c, 0), vertices (±a, 0)e = c/a (0 < e < 1), latus rectum = 2b²/a

The major axis has length 2a and the minor axis 2b. If the larger denominator is under y², as in x²/b² + y²/a² = 1, the major axis is along the y-axis and the foci are (0, ±c). The closer e is to 0, the rounder the ellipse; a circle is the limiting case e = 0. For JEE also note the directrices of x²/a² + y²/b² = 1, the lines x = ±a/e.

Worked example: For x²/25 + y²/9 = 1, find the foci, eccentricity and length of the latus rectum.
Solution: The larger denominator is under x², so a² = 25 and b² = 9: a = 5, b = 3. Then c = √(25 − 9) = 4. Foci (±4, 0); e = 4/5; latus rectum = 2 × 9/5 = 18/5.

Hyperbola

A hyperbola is the set of points in a plane the difference of whose distances from two fixed points (the foci) is a constant, taken as 2a in absolute value.

x²/a² − y²/b² = 1transverse axis along the x-axisc² = a² + b², foci (±c, 0), vertices (±a, 0)e = c/a (e > 1), latus rectum = 2b²/a

The transverse axis has length 2a and the conjugate axis 2b. For a hyperbola a need not be larger than b; what decides the orientation is which term is positive. In y²/a² − x²/b² = 1 the foci are (0, ±c). The lines y = ±(b/a)x are the asymptotes of x²/a² − y²/b² = 1, which the branches approach but never meet. When a = b, the hyperbola is called rectangular and e = √2.

Ellipse and hyperbola with their fociwww.iitmedicoguide.comEllipse: x²/a² + y²/b² = 1c² = a² − b², e = c/a < 1xyPF₁F₂−aab−bPF₁ + PF₂ = 2aHyperbola: x²/a² − y²/b² = 1c² = a² + b², e = c/a > 1xyy = (b/a)xPF₁F₂−aa|PF₁ − PF₂| = 2awww.iitmedicoguide.com
In the ellipse the foci lie inside the curve and the distances from any point add up to 2a. In the hyperbola the foci lie beyond the vertices and the distances differ by 2a; the dashed lines are the asymptotes.
Worked example: Find the foci, vertices, eccentricity and latus rectum of 9y² − 4x² = 36.
Solution: Divide by 36: y²/4 − x²/9 = 1. The positive term has y, so the transverse axis is vertical, with a² = 4 and b² = 9: a = 2, b = 3, c = √(4 + 9) = √13. Foci (0, ±√13); vertices (0, ±2); e = √13/2; latus rectum = 2 × 9/2 = 9.

Summary

 Parabola y² = 4axEllipse x²/a² + y²/b² = 1Hyperbola x²/a² − y²/b² = 1
Eccentricitye = 1e = c/a < 1e = c/a > 1
Relationonly one constant, ac² = a² − b²c² = a² + b²
Foci(a, 0)(±c, 0)(±c, 0)
Latus rectum4a2b²/a2b²/a
Common mistakes: (1) Using c² = a² + b² for an ellipse or c² = a² − b² for a hyperbola. (2) Assuming a is always under x²; in an ellipse a² is the larger denominator, in a hyperbola a² goes with the positive term. (3) For x² = −16y, writing the focus as (−4, 0) instead of (0, −4). (4) Forgetting to divide by the constant to bring the equation to standard form (the right side must be 1). (5) Taking the latus rectum of a parabola as 2a; it is 4a. (6) Not checking g² + f² − c > 0 before calling an equation a circle.

Exam focus

  • Centre and radius from the general equation of a circle; circle through given points or with given ends of a diameter.
  • Focus, directrix, axis and latus rectum for all four standard parabolas.
  • Foci, vertices, eccentricity and latus rectum of an ellipse or hyperbola, including vertical orientation.
  • Finding the equation of a conic from given foci, vertices or eccentricity.
  • Asymptotes and the rectangular hyperbola.

Practice questions

The centre and radius of x² + y² − 4x + 6y − 12 = 0 are:

  1. (2, −3), 5
  2. (−2, 3), 5
  3. (2, −3), √12
  4. (−2, 3), 25
Show answer
A. (x − 2)² + (y + 3)² = 12 + 4 + 9 = 25.

The focus of the parabola x² = −16y is:

  1. (0, 4)
  2. (0, −4)
  3. (−4, 0)
  4. (4, 0)
Show answer
B. 4a = 16 gives a = 4; it opens downward, so the focus is (0, −4).

The eccentricity of x²/25 + y²/16 = 1 is:

  1. 4/5
  2. 5/3
  3. 3/5
  4. 3/4
Show answer
C. c = √(25 − 16) = 3 and e = c/a = 3/5.

The length of the latus rectum of y² = 12x is:

  1. 3
  2. 6
  3. 4
  4. 12
Show answer
D. 4a = 12.

The foci of the hyperbola x²/16 − y²/9 = 1 are:

  1. (±√7, 0)
  2. (0, ±5)
  3. (±4, 0)
  4. (±5, 0)
Show answer
D. c² = 16 + 9 = 25. (±√7, 0) would come from wrongly subtracting.

A plane that does not pass through the vertex and is parallel to a generator of the cone cuts it in a:

  1. parabola
  2. ellipse
  3. hyperbola
  4. circle
Show answer
A. This is the case β = α.

The ellipse with vertices (±13, 0) and foci (±5, 0) is:

  1. x²/169 + y²/25 = 1
  2. x²/144 + y²/169 = 1
  3. x²/169 + y²/144 = 1
  4. x²/169 − y²/144 = 1
Show answer
C. a = 13, c = 5, so b² = 169 − 25 = 144.

The eccentricity of a rectangular hyperbola is:

  1. 1
  2. √2
  3. 2
  4. 1/√2
Show answer
B. a = b gives c = a√2, so e = √2.
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