NDA Maths · Conics

Hyperbola — Foci & Eccentricity

A hyperbola is the same family with e > 1; its standard form x²/a² − y²/b² = 1 has foci a distance c from the centre where c² = a² + b² — note the PLUS, the one sign that flips from the ellipse.

Why this matters

Only 4 PYQs, all direct: read a and b from the standard form, use c² = a² + b² (not minus), and the foci and eccentricity follow. The parametric form (a sec θ, b tan θ) appears occasionally.

Concept 1 of 2

Standard Form, Foci & Eccentricity

Intuition

The hyperbola mirrors the ellipse with one change: the foci sit FURTHER out than the vertices, so c² = a² + b². Normalise the equation to '= 1', read a and b, and everything follows.

Definition

For x2a2y2b2=1\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1:

  • Foci: (±c,0)(\pm c, 0) with c2=a2+b2c^2 = a^2 + b^2; distance between foci =2c= 2c.
  • Eccentricity: e=ca>1e = \dfrac{c}{a} > 1.
  • Asymptotes: y=±baxy = \pm \dfrac{b}{a}x.
  • Always normalise to =1=1 first: 25x275y2=22525x^2 - 75y^2 = 225 becomes x29y23=1\tfrac{x^2}{9} - \tfrac{y^2}{3} = 1.

Hyperbola foci & eccentricity

c2=a2+b2,e=ca>1c^2 = a^2 + b^2, \qquad e = \dfrac{c}{a} > 1

Worked example

Find the distance between the foci of x29y216=1\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1.
  1. a2=9, b2=16a^2 = 9,\ b^2 = 16, so c2=a2+b2=25c^2 = a^2 + b^2 = 25, c=5c = 5.
  2. Distance between foci =2c= 2c.
Answer:1010.

From the bank · past-year question

Example 1ConicsEASY
What is the distance between the foci of the hyperbola x24y2=1x^2-4y^2=1?

[Q65 · Sep · 2025]

Hyperbola uses PLUS: c² = a² + b²

The ellipse has c2=a2b2c^2 = a^2 - b^2; the hyperbola has c2=a2+b2c^2 = a^2 + b^2. Carrying the ellipse's minus sign into a hyperbola is the single most common slip in this subtopic.

Concept 2 of 2

Parametric Form & θ-Independent Properties

Intuition

A point given as (a sec θ, b tan θ) is on a hyperbola — recover the standard form using sec²θ − tan²θ = 1. Some hyperbola families have foci that don't move as a parameter changes, because c² stays constant.

Definition

  • Parametric point: (asecθ, btanθ)(a\sec\theta,\ b\tan\theta) lies on x2a2y2b2=1\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 (use sec2θtan2θ=1\sec^2\theta - \tan^2\theta = 1). A point like (3tanθ,2secθ)(3\tan\theta, 2\sec\theta) gives a hyperbola opening along yy.
  • **θ\theta-independent foci:** for x2cos2θy2sin2θ=1\dfrac{x^2}{\cos^2\theta} - \dfrac{y^2}{\sin^2\theta} = 1, c2=cos2θ+sin2θ=1c^2 = \cos^2\theta + \sin^2\theta = 1, so the foci are (±1,0)(\pm 1, 0) regardless of θ\theta, while e=secθe = \sec\theta.

Parametric identity

sec2θtan2θ=1\sec^2\theta - \tan^2\theta = 1

Worked example

A point on a curve is (2secθ,5tanθ)(2\sec\theta, \sqrt{5}\tan\theta). Find the eccentricity.
  1. Eliminate θ\theta: x24y25=sec2θtan2θ=1\dfrac{x^2}{4} - \dfrac{y^2}{5} = \sec^2\theta - \tan^2\theta = 1, so a2=4, b2=5a^2 = 4,\ b^2 = 5.
  2. c2=a2+b2=9c^2 = a^2 + b^2 = 9, c=3c = 3; e=ca=32e = \dfrac{c}{a} = \dfrac{3}{2}.
Answer:e=32e = \dfrac{3}{2}.

From the bank · past-year question

Example 2ConicsMODERATE
If any point on a hyperbola is (3tanθ,2secθ)(3\tan\theta, 2\sec\theta), then what is the eccentricity of the hyperbola?

[Q64 · Apr · 2021]

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Watch out for (1)

Mastery check — 2 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1ConicsEASY
What is the distance between the two foci of the hyperbola 25x275y2=22525x^2 - 75y^2 = 225?

[Q59 · Apr · 2025]

Example 2ConicsMODERATE
Consider hyperbola x2cos2θy2sin2θ=1\frac{x^2}{\cos^2\theta}-\frac{y^2}{\sin^2\theta}=1. (1) Two foci are independent of θ\theta. (2) Eccentricity is secθ\sec\theta. (3) Distance between foci is 2. How many statements are correct?

[Q86 · Sep · 2023]

Drill every past-year question on this subtopic

4 questions from the bank — paginated, with cart and Word-export support.