Conic Section Classification

On this page
  1. Direct answer
  2. What you must remember
  3. Completing the square before classifying
  4. How the chapter is examined
  5. Frequently asked questions
  6. Related topics

Direct answer

Eccentricity is the single number that separates the conic sections: the focus-directrix definition SP = e × PM gives a circle at e = 0, an ellipse for 0 < e < 1, a parabola at e = 1 and a hyperbola for e > 1, with e = √2 marking the rectangular hyperbola. The standard forms: y^2 = 4ax (parabola), x^2/a^2 + y^2/b^2 = 1 with b^2 = a^2(1 − e^2) (ellipse), x^2/a^2 − y^2/b^2 = 1 with b^2 = a^2(e^2 − 1) (hyperbola). From ax^2 + 2hxy + by^2 + ..., classify by the discriminant invariant: h^2 < ab an ellipse, h^2 = ab a parabola, h^2 > ab a hyperbola, a + b = 0 rectangular.

What you must remember

  • Definition: SP = e × PM, with S the focus, P a point on the curve and M the foot of the perpendicular from P to the directrix — every standard equation descends from it.
  • Parabola: y^2 = 4ax has focus (a, 0), directrix x = −a and latus rectum 4a; eccentricity exactly 1.
  • Ellipse: x^2/a^2 + y^2/b^2 = 1 with a > b carries foci (±ae, 0), b^2 = a^2(1 − e^2), and focal distances summing to 2a.
  • Hyperbola: x^2/a^2 − y^2/b^2 = 1 carries foci (±ae, 0), b^2 = a^2(e^2 − 1), and focal distances differing by 2a.
  • Rectangular hyperbola: a = b forces e = √2; xy = c^2 is the same creature with the coordinate axes as asymptotes.
  • Discriminant test: on ax^2 + 2hxy + by^2 + ..., compare h^2 with ab: less than for ellipse-type (circle when a = b, h = 0), equal for parabola-type, greater for hyperbola-type; a + b = 0 certifies rectangular.
  • Degenerates: an ellipse can collapse to a point and a "hyperbola" can be a line pair — the full equation, not just its quadratic part, decides.

Completing the square before classifying

Consider 9x^2 + 16y^2 = 144. Divide by 144: x^2/16 + y^2/9 = 1 — an ellipse with a = 4, b = 3. Then e = √(1 − b^2/a^2) = √(1 − 9/16) = √7/4, the foci are (±√7, 0) (since c = √(a^2 − b^2) = √7), and the directrices are x = ±a/e = ±16/√7. Now flip one sign: 9x^2 − 16y^2 = 144 becomes x^2/16 − y^2/9 = 1 — a hyperbola with e = √(1 + 9/16) = 5/4 and foci (±5, 0). Everything downstream — foci, directrices, latera recta — is read from the two denominators and the sign between the terms, which is why the classification step deserves its full minute: mislabel the curve and every subsequent answer inherits the error. For shifted conics like 4x^2 + 9y^2 − 16x + 18y + 11 = 0, complete squares first; the classification itself is translation-invariant.

How the chapter is examined

JEE Main tests direct classification and parameter read-offs: identify the conic, quote eccentricity, foci, directrix. JEE Advanced rotates the frame — literally: equations carrying an xy term are classified through h^2 versus ab, sometimes after a rotation whose angle satisfies tan 2α = 2h/(a − b). The recurring losses: calling the larger denominator b in an ellipse (a is always the semi-major axis, the larger one); quoting e = √2 for a hyperbola that is not rectangular; and overlooking degenerate cases, where the discriminant says "ellipse" but the equation has no real points. The identity xy = c^2 = a rectangular hyperbola with e = √2 is the single most quotable line in the chapter — several JEE Main sessions have asked nothing more.

Frequently asked questions

What eccentricity defines each conic type?

Circle e = 0, ellipse 0 < e < 1, parabola e = 1, hyperbola e > 1, rectangular hyperbola e = √2.

What is the eccentricity of x^2/16 + y^2/9 = 1?

√7/4, from e = √(1 − b^2/a^2) with a = 4, b = 3.

How does the discriminant test classify ax^2 + 2hxy + by^2?

Compare h^2 with ab: smaller means ellipse-type, equal means parabola-type, larger means hyperbola-type.

What is the eccentricity of a rectangular hyperbola?

Exactly √2, equivalent to a = b or to perpendicular asymptotes.

What curve is xy = 4?

A rectangular hyperbola referred to its asymptotes — the coordinate axes serve as them.

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