Skip to main content
Next exam on November 14 – 15, 2026:
All concepts

Conic Sections: Basic Quantities

圆锥曲线基本量

Standard equations of the ellipse, hyperbola and parabola, and converting between a, b, c, foci, directrix, asymptotes and eccentricity.

Topic drill: learn this step by step, with real exam questions and a mastery check
12 real-exam questions · 5 sittings43 in mock papers33 practice

Formulas

  • Ellipse: Standard Form

    Formula
    x2a2+y2b2=1\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1

    Applies when：a > b > 0 (foci on the x-axis); c² = a² − b²

  • Hyperbola Asymptotes

    Formula
    y=±baxy=\pm\frac{b}{a}x

    Applies when：For x²/a² − y²/b² = 1 (a>0, b>0); c² = a² + b²

  • Parabola Focus and Directrix

    Formula
    y2=2px (p>0):F(p2, 0),x=−p2y^{2}=2px\ (p>0):\quad F\left(\frac{p}{2},\,0\right),\quad x=-\frac{p}{2}
    p
    distance from focus to directrix (p > 0)

  • Eccentricity (Ellipse & Hyperbola)

    Definition
    e=cae=\frac{c}{a}

    Applies when：For standard-form ellipses and hyperbolas: ellipse 0 < e < 1 (c² = a² − b²); hyperbola e > 1 (c² = a² + b²)

    c
    linear eccentricity (half the focal distance)
    a
    semi-major axis (ellipse) / semi-transverse axis (hyperbola)

How the real exam tests it

Questions from real CSCA papers that test this concept. Work each one out before you open the answer.

  1. CSCA Math December 2025 Past Paper
    The foci of x2−y2=1x^2-y^2=1 are ( ).
    • A．(0,±2)(0,\pm\sqrt{2})
    • B．(±2,0)(\pm2,0)
    • C．(±2,0)(\pm\sqrt{2},0)
    • D．(0,±2)(0,\pm2)
    Show answer & solution

    Answer: C

    Answer: C For x2−y2=1x^2-y^2=1, a2=1a^2=1, b2=1b^2=1, so c2=2c^2=2. The foci are (±2,0)(\pm\sqrt{2},0). Choose C.
    See it in the full CSCA Math December 2025 Past Paper
  2. CSCA Math March 2026 Past Paper
    The equation of the directrix of the parabola y=−x2/4y = - x ^ { 2 } / 4 is
    • A．x=1/16x = 1 / 16
    • B．y=1/16y = 1 / 16
    • C．y=1y = 1
    • D．x=1x = 1
    Show answer & solution

    Answer: C

    Rewrite as x2=−4yx ^ { 2 } = - 4 y. Standard form x2=−2px ^ { 2 } = - 2 p gives 2p=4,p=22 p = 4 , p = 2. Directrix: y=p/2=1y = p / 2 = 1.
    Take the full CSCA Math March 2026 Past Paper — timed & graded
  3. CSCA Math April 2026 Past Paper
    For a hyperbola with focal length 4 and eccentricity e=2e = 2, find aa.
    • A．2
    • B．62\frac { \sqrt { 6 } } { 2 }
    • C．52\frac { \sqrt { 5 } } { 2 }
    • D．1
    Show answer & solution

    Answer: D

    Focal length 2c=4⇒c=22 c = 4 \Rightarrow c = 2. From e=ca=2:a=c2=1e = \frac { c } { a } = 2 : a = \frac { c } { 2 } = 1.
    Take the full CSCA Math April 2026 Past Paper — timed & graded

Real-exam evidence

Recorded past-paper appearances are listed below; online verification is labelled separately.

12 questions across 5 sittings1 of 5 sittings available to verify free online

43 questions across 15 mock papers — our own papers, so a coverage signal rather than real-exam evidence

33 related practice questions availablePractise free: Analytic Geometry

Sitting evidence comes from the CSCA past papers on record; questions whose original paper is missing are excluded, and sprint papers are deliberately not counted.

Glossary

ellipsehyperbolaparabolaeccentricity

Related concepts

Vectors and the Dot ProductQuadrants of the Coordinate PlaneEquations of LinesEquations of CirclesSpace Rectangular Coordinates