Circles and conic sections

NUST NET (Engineering) · Maths · Analytic geometry. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.

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The idea

Each of these curves has a standard equation, and the test asks you to read its features straight from that equation. So the first step is always to put the equation in standard form, completing the square if x or y terms are present.

For a circle you read the centre and radius. For a parabola you find a from 4a, then the focus is a units inside the curve from the vertex and the directrix is a units outside. For an ellipse or hyperbola you find c, the distance from the centre to each focus, and the eccentricity e = c/a.

The eccentricity names the curve: 0 for a circle, less than 1 for an ellipse, exactly 1 for a parabola, more than 1 for a hyperbola.

Rules to remember

Common mistake

Mixing up the two formulas for c: the ellipse uses c2 = a2 − b2 but the hyperbola uses c2 = a2 + b2. Check your answer against e < 1 for an ellipse and e > 1 for a hyperbola.

Worked example

Find the focus of the parabola x2 − 4x − 8y + 12 = 0.

  1. Complete the square: (x − 2)2 − 4 = 8y − 12, so (x − 2)2 = 8(y − 1).
  2. Compare with X2 = 4aY: vertex (2, 1) and 4a = 8, so a = 2.
  3. The parabola opens upwards, so the focus is 2 units above the vertex: (2, 1 + 2).

Answer: Focus (2, 3); the directrix is y = −1.

Practice questions

Try each one, then open the answer.

1. The vertex of the parabola y2 − 4y − 8x + 12 = 0 is

  1. A
    (2, 1)
  2. B
    (1, 2)
  3. C
    (−1, 2)
  4. D
    (1, −2)
Show answer

Answer: B. Complete the square: y2 − 4y + 4 = 8x − 8, i.e. (y − 2)2 = 8(x − 1), so the vertex is (1, 2).

2. The circle with centre (2, −3) that passes through the origin is

  1. A
    x2 + y2 + 4x − 6y = 0
  2. B
    x2 + y2 − 4x + 6y − 13 = 0
  3. C
    x2 + y2 − 2x + 3y = 0
  4. D
    x2 + y2 − 4x + 6y = 0
Show answer

Answer: D. r2 = 22 + 32 = 13, so (x − 2)2 + (y + 3)2 = 13, which expands to x2 + y2 − 4x + 6y = 0. A circle through the origin has no constant term.

3. The directrices of the ellipse x2/25 + y2/9 = 1 are

  1. A
    x = ±25/4
  2. B
    x = ±4
  3. C
    y = ±25/4
  4. D
    x = ±15/4
Show answer

Answer: A. c = √(25 − 9) = 4, so e = 4/5 and the directrices are x = ±a/e = ±5 ÷ (4/5) = ±25/4. The lines x = ±4 pass through the foci.

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