∫ SJ Maths Mathematics Notes
UP PGT Tracker UP TGT Tracker
4. Coordinate Geometry of Two Dimensions

Classification as ellipse

This section covers the identification and classification of the general second-degree equation as an ellipse.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade
📖 Topic progress
○ Mark as complete

Definition and statement

General Second-Degree Equation Ellipse Classification Principle
The general second-degree equation in two variables given by $$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$$ represents a real non-degenerate ellipse if and only if the discriminant $\Delta \neq 0$ and $h^2 < ab$, where $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}$. If $\Delta = 0$, it degenerates to a single point (point ellipse).

📐 Geometric Interpretation

A non-degenerate ellipse represents a closed, bounded, central conic section whose plane intersects a double-napped right circular cone at an angle oblique to the axis, completely enclosing a bounded region. The center of the conic is given by the intersection of the partial derivative lines $\frac{\partial S}{\partial x} = 0$ and $\frac{\partial S}{\partial y} = 0$.

⚙️ Algebraic & Structural Meaning

The condition $h^2 < ab$ guarantees that the quadratic form $ax^2 + 2hxy + by^2$ is definite (either positive or negative definite depending on $c$ and $\Delta$), ensuring that the cross-product term can be eliminated via a rotation of coordinate axes by an angle $\theta$ given by $\tan 2\theta = \frac{2h}{a - b}$.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Non-zero Determinant (Non-degeneracy) \Delta \neq 0 Strictly Mandatory / Necessary If $\Delta = 0$, the conic degenerates into intersecting lines, parallel lines, a single point, or no locus at all, failing to form a true 2D areal ellipse.
Angular Discriminant Inequality h^2 < ab Strictly Mandatory If $h^2 = ab$, the curve classifies as a parabola; if $h^2 > ab$, it classifies as a hyperbola.
Sign Compatibility of Discriminant and Leading Coefficients \Delta \cdot (a + b) < 0 \text{ (for real ellipse)} Essential for real locus If $\Delta \cdot (a + b) \ge 0$ while $h^2 < ab$, the equation represents an imaginary ellipse (empty set in $\mathbb{R}^2$).

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Quick Ellipse vs Parabola vs Hyperbola Test

$$h^2 - ab \begin{cases} < 0 & \text{Ellipse} \\ = 0 & \text{Parabola} \\ > 0 & \text{Hyperbola} \end{cases}$$

Rule: Instantly check the sign of $h^2 - ab$ after extracting coefficients from $ax^2 + 2hxy + by^2 + \dots = 0$.

Application: undefined
02

Center Coordinates Formula

$$(x_0, y_0) = \left( \frac{hf - bg}{ab - h^2}, \frac{hg - af}{ab - h^2} \right)$$

Rule: Obtain the center instantly by solving $\frac{\partial S}{\partial x} = 0$ and $\frac{\partial S}{\partial y} = 0$ using Cramer's rule.

Application: Frequently asked in KVS PGT and DSSSB for finding the center of a rotated ellipse.
03

Sum of Inverse Squared Semi-Axes

$$\frac{1}{a_1^2} + \frac{1}{b_1^2} = \frac{a+b}{...} \text{ (via invariants)}$$ invariant relation under rotation: $a + b = \frac{1}{a_1^2} + \frac{1}{b_1^2}$ (when shifted to center)

Rule: Use trace invariance $a + b = \lambda_1 + \lambda_2$ where $\lambda_1, \lambda_2$ are roots of the characteristic equation $(\lambda - a)(\lambda - b) = h^2$.

Application: Used in advanced UP PGT problems to find lengths of semi-axes without full rotation.
04

Condition for Circular Locus (Special Ellipse)

$$a = b \text{ and } h = 0$$

Rule: If $a = b$ and $h = 0$, the ellipse reduces to a circle (eccentricity $e = 0$).

Application: Quick identification of circles disguised as general second-degree equations.
05

Determinant Quick Expansion for $\Delta$

$$\Delta = a(bc - f^2) - h(hc - fg) + g(hf - bg)$$

Rule: Expand along the first row to check for non-degeneracy.

06

Elimination via Orthogonal Axes Property

$$\tan 2\theta = \frac{2h}{a - b}$$

Rule: If $a = b$, the axes are inclined at $\theta = \frac{\pi}{4}$, immediately eliminating options with arbitrary rotation angles.

Application: Rapid elimination trick for matching equations with rotated conics.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Forgetting the $2$ in $2hxy$ and $2gx$ and $2fy$

Common Pitfall: Students take the coefficient of $xy$ directly as $h$ instead of dividing by $2$. For example, in $3x^2 + 10xy + 3y^2$, $2h = 10 \implies h = 5$, not $10$.

Counterexample: Equation: $3x^2 + 10xy + 3y^2 = 0 \implies h = 5, a = 3, b = 3 \implies h^2 = 25 > 9$ (Hyperbola, not Ellipse!)

Why it fails: Using $h = 10$ gives $100 < 9$ (false positive for ellipse), leading to incorrect option selection.

🛡️ Exam Defense: Always match the given equation with the standard template $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ and explicitly write $h = \frac{\text{coeff of } xy}{2}$.
⚠️ COMMON ERROR #02

Confusing Real vs. Imaginary Ellipse

Common Pitfall: Checking only $h^2 < ab$ and $\Delta \neq 0$, and concluding it's an ellipse without checking if $\Delta(a+b) < 0$.

Counterexample: Equation: $x^2 + 2xy + 2y^2 + 2x + 2y + 5 = 0$

Why it fails: It satisfies $h^2 < ab$ ($1^2 < 1 \cdot 2$), but yields an imaginary locus (empty set) because $\Delta > 0$ with correct sign balance failing.

🛡️ Exam Defense: Always evaluate the full determinant $\Delta$ and ensure the constant term after shifting to center is negative for a real ellipse.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Ellipse\Delta \neq 0, \; h^2 < ab\Deltaabc + 2fgh - af^2 - bg^2 - ch^2 \neq 0Closed bounded convex oval curveCoefficients of $x^2$ and $y^2$ have same sign, $h^2 < ab$
Parabola\Delta \neq 0, \; h^2 = abh^2 = abOpen curve with a single branch extending to infinityDiscriminant of quadratic part is zero ($h^2 = ab$)
Hyperbola\Delta \neq 0, \; h^2 > abh^2 > abOpen curve with two separate branches and asymptotesCoefficients of $x^2$ and $y^2$ have opposite signs or large $h$
Point Ellipse\Delta = 0, \; h^2 < ab\Delta = 0Degenerates to a single isolated pointDiscriminant $\Delta$ vanishes identically while $h^2 < ab$

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the nature of the conic represented by the equation $2x^2 - 5xy + 3y^2 - 2x - 3y + 2 = 0$.

(A) Ellipse(B) Parabola(C) Hyperbola(D) Pair of straight lines
✓ Correct Answer: (C)
📚 Method A: Formal Textbook Derivation
Step 1: Identify coefficients: $a = 2, 2h = -5 \implies h = -5/2, b = 3, g = -1, f = -3/2, c = 2$. Step 2: Calculate $h^2 - ab = \left(-\frac{5}{2}\right)^2 - (2)(3) = \frac{25}{4} - 6 = \frac{1}{4} > 0$. Step 3: Since $h^2 > ab$, the conic is a hyperbola.
⚡ Method B: 25-Second Exam Shortcut
Compute $h^2 - ab = (-2.5)^2 - (2)(3) = 6.25 - 6 = +0.25 > 0$. Positive means hyperbola instantly.
💡 Tip: Do not waste time calculating $\Delta$ if $h^2 - ab > 0$ and coefficients are non-zero; it is definitively a hyperbola.
CANONICAL PROBLEM 02 Dual-Method Breakdown

For what value of $\lambda$ does the equation $x^2 + 4xy + \lambda y^2 - 2x + 4y + 1 = 0$ represent an ellipse?

(A) $\lambda > 4$(B) $\lambda < 2$(C) $\lambda > 2$(D) $\lambda = 4$
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Identify $a = 1, 2h = 4 \implies h = 2, b = \lambda$. Step 2: The condition for an ellipse is $h^2 < ab$. Step 3: $2^2 < (1)(\lambda) \implies 4 < \lambda \implies \lambda > 4$.
⚡ Method B: 25-Second Exam Shortcut
Set $h^2 < ab \implies 2^2 < 1 \cdot \lambda \implies \lambda > 4$. Mark option (A) in 10 seconds.
💡 Tip: Watch out for inequality direction changes when dividing by negative numbers (not applicable here, but vital in general).

Practice questions

Choose an answer to check it and see the solution:

Q1
The equation $7x^2 - 8xy + y^2 + 14x - 2y + 2 = 0$ represents:
A An ellipse
B A hyperbola
C A parabola
D A pair of straight lines
Derivation & Analysis:

Here $a = 7, h = -4, b = 1$. Then $h^2 - ab = (-4)^2 - (7)(1) = 16 - 7 = 9 > 0$. Since $h^2 > ab$, the conic is a hyperbola.

Q2
If the general second-degree equation represents a real ellipse, then the determinant $\Delta$ and $a+b$ must satisfy:
A $\Delta$ = 0, a+b $\neq$ 0
B $\Delta$ $\neq$ 0, $\Delta$(a+b) < 0
C $\Delta$ $\neq$ 0, $\Delta$(a+b) > 0
D $\Delta = 0, h^2 = ab$
Derivation & Analysis:

For a non-degenerate real ellipse, $\Delta \neq 0$ and the product of $\Delta$ and $(a+b)$ must be strictly negative ($\Delta(a+b) < 0$).

Q3
The equation $x^2 - 2xy + 2y^2 - 2x + 2y + 1 = 0$ represents:
A A real ellipse
B An imaginary ellipse
C A parabola
D A hyperbola
Derivation & Analysis:

Here $a = 1, h = -1, b = 2$. $h^2 - ab = 1 - 2 = -1 < 0$ (Ellipse). Calculating $\Delta$, we find $\Delta \neq 0$ and real locus conditions are satisfied.

Q4
The center of the ellipse given by $2x^2 + 3y^2 - 4x - 6y - 1 = 0$ is:
A (1, 1)
B (1, 2)
C (1, 3)
D (2, 1)
Derivation & Analysis:

Differentiating partially w.r.t $x$: $4x - 4 = 0 \implies x = 1$. Differentiating w.r.t $y$: $6y - 6 = 0 \implies y = 1$. Center is $(1, 1)$.

Q5
If $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a point ellipse, then:
A $\Delta = 0 \text{ and } h^2 < ab$
B $\Delta \neq 0 \text{ and } h^2 < ab$
C $\Delta = 0 \text{ and } h^2 = ab$
D $\Delta \neq 0 \text{ and } h^2 > ab$
Derivation & Analysis:

A point ellipse is a degenerate conic where $\Delta = 0$ while the quadratic part remains elliptic ($h^2 < ab$).

Q6
The condition that the equation $x^2 + kxy + y^2 - 5x - 5y + 6 = 0$ represents an ellipse is:
A $-2 < k < 2$
B $k > 2$
C $k < -2$
D $k \neq \pm 2$
Derivation & Analysis:

Here $a = 1, b = 1, 2h = k \implies h = k/2$. For ellipse, $h^2 < ab \implies \left(\frac{k}{2}\right)^2 < 1 \cdot 1 \implies \frac{k^2}{4} < 1 \implies k^2 < 4 \implies -2 < k < 2$.

Q7
What is the angle of rotation $\theta$ to eliminate the $xy$ term in $ax^2 + 2hxy + by^2 = 0$ when $a \neq b$?
A $\tan \theta = \frac{2h}{a-b}$
B $\tan 2\theta = \frac{2h}{a-b}$
C $\tan 2\theta = \frac{h}{a-b}$
D $\tan \theta = \frac{h}{a-b}$
Derivation & Analysis:

Standard transformation formula for eliminating the cross-product term is $\tan 2\theta = \frac{2h}{a-b}$.

Q8
If $a = b$ in the general second-degree equation and $h = 0$, the conic is:
A A hyperbola
B A parabola
C A circle
D A pair of lines
Derivation & Analysis:

When $a = b$ and $h = 0$ with $g, f, c$ appropriately scaled, the equation reduces to $x^2 + y^2 + 2gx + 2fy + c = 0$, which is a circle.

Q9
The sum of the coefficients of $x^2$ and $y^2$ in a rotated conic is invariant under coordinate rotation. For $ax^2 + 2hxy + by^2$, this invariant is:
A a - b
B a + b
C $ab - h^2$
D $a^2$ + $b^2$
Derivation & Analysis:

The trace $a + b$ and the discriminant $h^2 - ab$ are fundamental invariants under orthogonal coordinate rotations.

Q10
For the conic $4x^2 + 9y^2 - 8x - 36y + 4 = 0$, the eccentricity $e$ is:
A $\frac{\sqrt{5}}{3}$
B $\frac{2}{3}$
C $\frac{1}{3}$
D $\frac{\sqrt{5}}{2}$
Derivation & Analysis:

Rewriting in standard form: $\frac{(x-1)^2}{9} + \frac{(y-2)^2}{4} = 1$. Here $a^2 = 9, b^2 = 4$ ($a > b$). Eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3}$.

Common questions

Examiners deliberately provide equations with $2hxy$ written as a combined coefficient (e.g., $6xy$). Candidates often forget to divide by $2$ to find $h$, leading to $h^2 > ab$ errors.

A standard real ellipse has $\Delta \neq 0$. If $\Delta = 0$ while $h^2 < ab$, the equation represents a degenerate conic collapsing into a single real point.

Yes, memorizing $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2$ or the 3x3 determinant form is essential for solving non-degeneracy questions within the 50-second time limit per question.

← Previous Topic
Classification as circle
Next Topic →
Classification as hyperbola