Conic Sections: Edge Cases and Subtle Traps (7)

easy • 4 min read
Tags Conic Sections

Question

Identify the conic section represented by 4x2−9y2−16x−18y−29=04x^2 - 9y^2 - 16x - 18y - 29 = 0, find its centre, foci, and eccentricity. Then explain which subtle JEE trap appears in this kind of “complete-the-square” question.

Solution — Step by Step

4x2−16x−9y2−18y=294x^2 - 16x - 9y^2 - 18y = 29

4(x2−4x)−9(y2+2y)=294(x^2 - 4x) - 9(y^2 + 2y) = 29

Complete the squares:

  • x2−4x=(x−2)2−4x^2 - 4x = (x - 2)^2 - 4
  • y2+2y=(y+1)2−1y^2 + 2y = (y + 1)^2 - 1

Substitute:

4[(x−2)2−4]−9[(y+1)2−1]=294[(x-2)^2 - 4] - 9[(y+1)^2 - 1] = 29

4(x−2)2−9(y+1)2=29+16−9=364(x-2)^2 - 9(y+1)^2 = 29 + 16 - 9 = 36

Divide by 3636:

(x−2)29−(y+1)24=1\frac{(x-2)^2}{9} - \frac{(y+1)^2}{4} = 1

This is a horizontal hyperbola with centre (2,−1)(2, -1), a2=9a^2 = 9, b2=4b^2 = 4.

For a hyperbola: c2=a2+b2=9+4=13c^2 = a^2 + b^2 = 9 + 4 = 13, so c=13c = \sqrt{13}.

Eccentricity: e=c/a=13/3e = c/a = \sqrt{13}/3.

Foci: (2±13,−1)(2 \pm \sqrt{13}, -1).

The subtle trap: the signs of the coefficients of x2x^2 and y2y^2 are opposite — this is what makes it a hyperbola, not an ellipse. Many students reflexively apply the ellipse procedure when they see “conic section” and “complete the square”, missing the sign of the y2y^2 coefficient.

The second trap: in the hyperbola c2=a2+b2c^2 = a^2 + b^2, add. In the ellipse c2=a2−b2c^2 = a^2 - b^2, subtract. Mixing these up is the most common JEE conic error.

Why This Works

Every general conic Ax2+Cy2+Dx+Ey+F=0Ax^2 + Cy^2 + Dx + Ey + F = 0 (no xyxy term) reduces to a standard form by completing the square. The discriminant of Ax2+Cy2Ax^2 + Cy^2 tells you the type:

  • AC>0AC > 0 → ellipse (or circle if A=CA = C)
  • AC<0AC < 0 → hyperbola
  • A=0A = 0 or C=0C = 0 → parabola

In our problem, A=4A = 4, C=−9C = -9, so AC=−36<0AC = -36 < 0 — hyperbola.

Alternative Method

A faster check: rearrange to make one squared term equal 11:

4(x−2)236−9(y+1)236=1  ⟹  (x−2)29−(y+1)24=1\frac{4(x-2)^2}{36} - \frac{9(y+1)^2}{36} = 1 \implies \frac{(x-2)^2}{9} - \frac{(y+1)^2}{4} = 1

The minus sign between the two fractions is the giveaway: hyperbola. If both fractions had been added, ellipse.

For JEE Main shortcuts: any equation of the form X2p−Y2q=1\dfrac{X^2}{p} - \dfrac{Y^2}{q} = 1 with p,q>0p, q > 0 is a hyperbola of the form XX-axis transverse. The vertices are (±p,0)(\pm\sqrt{p}, 0), asymptotes are Y=±q/p XY = \pm\sqrt{q/p}\, X. Memorise these for speed.

Common Mistake

The single biggest trap on conic-section problems with “complete the square”:

  1. Sign of the constant after rearranging. When you bring the −29-29 to the right and absorb the −16-16 and +9+9 from completing the squares, the sign of the resulting constant determines whether you have a hyperbola or its conjugate. If the constant came out negative, divide both sides by it (with sign) to get 11 on the right — this can flip a horizontal hyperbola into a vertical one.

  2. Forgetting to multiply through by the leading coefficient before completing the square. For 4x2−16x4x^2 - 16x, factor out the 44 first: 4(x2−4x)4(x^2 - 4x), then complete inside the bracket. Trying to complete the square directly on 4x2−16x4x^2 - 16x without factoring gives 4(x−2)2−164(x-2)^2 - 16, which is correct — but students often forget to balance both sides when expanding back.

Final answer: Hyperbola, centre (2,−1)(2,-1), a=3a = 3, b=2b = 2, e=13/3e = \sqrt{13}/3, foci (2±13,−1)(2 \pm \sqrt{13}, -1).

Want to master this topic?

Read the complete guide with more examples and exam tips.

Go to full topic guide →

Try These Next

Find the area of triangle formed by tangent to ellipse and coordinate axes

medium

Conic Sections: Application Problems (5)

medium

Conic Sections: Common Mistakes and Fixes (1)

easy

Conic Sections: Conceptual Doubts Cleared (12)

hard

Conic Sections: Diagram-Based Questions (9)

hard