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4. Coordinate Geometry of Two Dimensions

General equation of second degree in two variables

The general equation of second degree in two variables, $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$, represents conic sections. This topic includes the discriminant condition $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$ or $\neq 0$ and homogenization techniques.

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Definition and statement

Discriminant Condition for Conics and Pair of Straight Lines
The general equation of second degree $$S(x, y) = ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$$ represents a degenerate conic (pair of straight lines) if and only if the determinant of the coefficients vanishes: $$\Delta = \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$$. If $\Delta \neq 0$, it represents a non-degenerate conic section (ellipse, parabola, hyperbola, or circle).

📐 Geometric Interpretation

Geometrically, when $\Delta = 0$, the second-degree curve splits into two distinct, coincident, or imaginary intersecting straight lines. When $\Delta \neq 0$, the equation traces a unique conic section whose eccentricity $e$ is dictated by the invariant invariants $I = a+b$ and $\Delta$.

⚙️ Algebraic & Structural Meaning

Algebraically, the condition $\Delta = 0$ is the exact resultant condition ensuring that the quadratic polynomial in two variables can be factored into two linear factors with real or complex coefficients.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Non-triviality of Second-Degree Terms N/A Strictly Mandatory If $a = h = b = 0$, the equation degenerates into a linear equation representing a single straight line, violating the degree condition.
Vanishing Determinant ($\Delta = 0$) for Pair of Lines N/A Strictly Mandatory If $\Delta \neq 0$, the locus can never be a pair of straight lines; it represents a proper conic section.
Condition for Real and Distinct Lines N/A Essential for real intersection If $h^2 < ab$ and $\Delta = 0$, the pair of straight lines is imaginary with a real point of intersection (the vertex).

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Angle Between Pair of Straight Lines

$$ \tan \theta = \pm \frac{2\sqrt{h^2 - ab}}{a + b} $$

Rule: Instantly find the acute angle between lines represented by $ax^2 + 2hxy + by^2 = 0$. For perpendicular lines, $a+b=0$. For coincident lines, $h^2=ab$.

Application: undefined
02

Point of Intersection of Pair of Lines

$$ \left( \frac{hf - bg}{ab - h^2}, \frac{gh - af}{ab - h^2} \right) $$

Rule: Obtained by partially differentiating $S(x,y)$ with respect to $x$ and $y$ and solving linear equations: $\frac{\partial S}{\partial x} = 0$ and $\frac{\partial S}{\partial y} = 0$.

03

Nature of Conic via Discriminant $h^2 - ab$

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

Rule: Examine the second-degree leading terms coefficient $h^2 - ab$ along with $a+b$ (for circle: $a=b, h=0$) to classify conics instantly.

Application: Frequent 1-mark identification questions in DSSSB and NVS PGT.
04

Combined Equation of Lines Joining Origin to Intersections

$$ ax^2 + 2hxy + by^2 + 2gx\left(\frac{lx+my}{1}\right) + 2fy\left(\frac{lx+my}{1}\right) + c\left(\frac{lx+my}{1}\right)^2 = 0 $$

Rule: Homogenize the second-degree equation using a linear equation $lx + my = 1$ to find lines joining the origin to the intersection points with a conic.

05

Length of Intercepts on Axes

$$ \text{X-intercept} = 2\frac{\sqrt{g^2 - ac}}{a}, \quad \text{Y-intercept} = 2\frac{\sqrt{f^2 - bc}}{b} $$

Rule: Set $y=0$ and $x=0$ respectively in the general conic equation to derive axis intercept lengths without full factorization.

06

Combined Equation of Pair of Perpendicular Lines

$$ (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \text{ with } a+b=0 $$

Rule: Eliminate cross-product term coefficient $h=0$ and ensure $a+b=0$ for orthogonal line pairs passing through given vertices.

Application: Rapid elimination trick for options in UP PGT.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Assuming $\Delta = 0$ Always Implies Real Lines

Common Pitfall: Candidates verify $\Delta = 0$ and immediately conclude the lines are real and distinct, forgetting to check $h^2 \ge ab$.

Counterexample: x^2 + y^2 = 0 \implies x^2 + y^2 + 0xy + 0x + 0y + 0 = 0

Why it fails: Here $\Delta = 0$, but $h^2 - ab = 0 - (1)(1) = -1 < 0$, representing a single real point (origin), not real intersecting lines.

🛡️ Exam Defense: Always check $h^2 - ab$ after verifying $\Delta = 0$ to distinguish between real intersecting lines, coincident lines, and point conics.
⚠️ COMMON ERROR #02

Misidentifying Parabola vs. Ellipse when $h^2 = ab$

Common Pitfall: Applying the $h^2 - ab = 0$ rule without ensuring $\Delta \neq 0$. If $\Delta = 0$, $h^2 = ab$ gives parallel or coincident lines, not a parabola.

Counterexample: x^2 + 2xy + y^2 + 2x + 2y + 1 = 0

Why it fails: Here $a=1, h=1, b=1 \implies h^2-ab=0$, but $\Delta = 0$, which factors into parallel lines $(x+y+1)^2 = 0$, not a parabola.

🛡️ Exam Defense: Always compute the full determinant $\Delta$ first before testing $h^2 - ab$ for conic classification.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Pair of Straight Lines\Delta = 0abc + 2fgh - af^2 - bg^2 - ch^2 = 0Degenerate conic splitting into two linear factorsEquation factorizable into linear terms
Parabola\Delta \neq 0, \; h^2 = abe = 1Open curve with eccentricity 1Square terms match cross-product condition
Ellipse\Delta \neq 0, \; h^2 < abe < 1Closed bounded symmetric oval shapeCoefficients of $x^2$ and $y^2$ same sign, $h^2 < ab$
Hyperbola\Delta \neq 0, \; h^2 > abe > 1Two unbounded open branches with asymptotesCoefficients of $x^2$ and $y^2$ opposite sign or $h^2 > ab$

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the value of $\lambda$ for which the equation $12x^2 - 10xy + 2y^2 + 11x - 5y + \lambda = 0$ represents a pair of straight lines.

(A) 2(B) -2(C) 0(D) 1
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Identify coefficients: $a = 12, 2h = -10 \implies h = -5, b = 2, 2g = 11 \implies g = 11/2, 2f = -5 \implies f = -5/2, c = \lambda$. Apply $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$. Substitute values to get $12(2)(\lambda) + 2(-5/2)(11/2)(-5) - 12(-5/2)^2 - 2(11/2)^2 - \lambda(-5)^2 = 0$, yielding $\lambda = 2$.
⚡ Method B: 25-Second Exam Shortcut
Use determinant expansion directly: $\begin{vmatrix} 12 & -5 & 11/2 \\ -5 & 2 & -5/2 \\ 11/2 & -5/2 & \lambda \end{vmatrix} = 0$. Solving gives $\lambda = 2$.
💡 Tip: Be extremely careful with fractions when dividing $2g, 2f, 2h$ to find $g, f, h$.
CANONICAL PROBLEM 02 Dual-Method Breakdown

The angle between the pair of lines given by $x^2 - 4xy + y^2 = 0$ is:

(A) \frac{\pi}{6}(B) \frac{\pi}{4}(C) \frac{\pi}{3}(D) \frac{\pi}{2}
✓ Correct Answer: (C)
📚 Method A: Formal Textbook Derivation
Here $a = 1, 2h = -4 \implies h = -2, b = 1$. Using the formula $\tan \theta = \pm \frac{2\sqrt{h^2 - ab}}{a + b}$, we get $\tan \theta = \frac{2\sqrt{(-2)^2 - (1)(1)}}{1 + 1} = \frac{2\sqrt{3}}{2} = \sqrt{3}$. Thus $\theta = \frac{\pi}{3}$.
⚡ Method B: 25-Second Exam Shortcut
Recognize standard form $ax^2 + 2hxy + by^2 = 0$. $\tan \theta = \frac{2\sqrt{4 - 1}}{2} = \sqrt{3} \implies \theta = 60^{\circ} = \frac{\pi}{3}$.
💡 Tip: Directly memorize standard angle results for symmetric equations like $x^2 - 4xy + y^2 = 0$.

Practice questions

Choose an answer to check it and see the solution:

Q1
The general equation of second degree $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a circle if:
A $a = b$ and $h = 0$
B $a = b$ and h $\neq$ 0
C $a + b = 0$ and $h = 0$
D $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$
Derivation & Analysis:

A circle requires second-degree coefficients of $x^2$ and $y^2$ to be equal ($a=b$) and the rectangle/cross-product term absent ($h=0$).

Q2
If the equation $ax^2 + 2hxy + by^2 = 0$ represents two coincident lines, then:
A $h^2 < ab$
B $h^2 = ab$
C $h^2 > ab$
D $a + b = 0$
Derivation & Analysis:

For coincident lines, the discriminant of the quadratic factorisation vanishes, giving $h^2 = ab$.

Q3
The point of intersection of the lines represented by $2x^2 + 5xy + 2y^2 + 5x + 5y + 2 = 0$ is:
A (1, 1)
B (-1, -1)
C (0, 0)
D (1, -1)
Derivation & Analysis:

Applying partial derivative formulas $\frac{\partial S}{\partial x} = 4x + 5y + 5 = 0$ and $\frac{\partial S}{\partial y} = 5x + 4y + 5 = 0$, solving yields $(-1, -1)$.

Q4
What conic does $xy = c^2$ represent?
A Circle
B Parabola
C Rectangular Hyperbola
D Ellipse
Derivation & Analysis:

In $xy = c^2$, $a=0, b=0, 2h=1 \implies h=1/2$, so $h^2 - ab = 1/4 > 0$ with $a+b=0$, representing a rectangular hyperbola.

Q5
If the lines $ax^2 + 2hxy + by^2 = 0$ are perpendicular to each other, then:
A $a = b$
B $a + b = 0$
C $h = 0$
D $h^2 = ab$
Derivation & Analysis:

The angle formula is $\tan \theta = \pm \frac{2\sqrt{h^2 - ab}}{a+b}$. For $\theta = \pi/2$, $\cot(\pi/2) = 0 \implies a+b = 0$.

Q6
The condition that one of the lines given by $ax^2 + 2hxy + by^2 = 0$ may coincide with $lx + my = 0$ is:
A $al^2 + 2hlm + bm^2 = 0$
B $am^2 - 2hlm + bl^2 = 0$
C $al^2 - 2hlm + bm^2 = 0$
D $bl^2 + 2hlm + am^2 = 0$
Derivation & Analysis:

Substituting $x = -my/l$ into the homogeneous equation yields the auxiliary condition $al^2 + 2hlm + bm^2 = 0$.

Q7
The equation $x^2 - 5xy + 6y^2 = 0$ represents:
A Real and distinct lines
B Real and coincident lines
C Imaginary lines with real intersection
D A circle
Derivation & Analysis:

Factoring gives $(x - 2y)(x - 3y) = 0$, which represents two real and distinct intersecting lines passing through the origin.

Q8
If $\Delta \neq 0$ and $h^2 - ab > 0$, the general second-degree equation represents a:
A Parabola
B Ellipse
C Hyperbola
D Circle
Derivation & Analysis:

By definition, when $\Delta \neq 0$ and $h^2 > ab$, the conic is a hyperbola.

Q9
The combined equation of the coordinate axes is:
A xy = 0
B $x + y = 0$
C $x^2 - y^2 = 0$
D $x^2 + y^2 = 0$
Derivation & Analysis:

The x-axis is $y=0$ and the y-axis is $x=0$. Their combined equation is $xy = 0$.

Q10
The distance between the parallel lines represented by $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ when $h^2 = ab$ and $bg^2 = af^2$ is:
A $2\sqrt{\frac{g^2 - ac}{a(a+b)}}$
B $\sqrt{\frac{g^2 - ac}{a}}$
C $2\sqrt{\frac{f^2 + g^2}{a+b}}$
D 0
Derivation & Analysis:

Using standard parallel line distance formula for conics when $\Delta = 0$ and $h^2 = ab$, the distance is $2\sqrt{\frac{g^2 - ac}{a(a+b)}}$.

Common questions

The vanishing of determinant $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$ is the universal algebraic test for degeneracy. Examiners frequently embed unknown parameters like $\lambda$ or $k$ inside the second-degree equation and ask candidates to find values for which it represents straight lines.

Always compute $h^2 - ab$ first (assuming $\Delta \neq 0$). If $h^2 - ab < 0$, it's an ellipse; if $h^2 - ab = 0$, it's a parabola; if $h^2 - ab > 0$, it's a hyperbola. This 5-second check bypasses lengthy coordinate transformations.

Never factorize cumbersome expressions during competitive exams. Instead, take partial derivatives with respect to $x$ and $y$ separately, set them to zero ($\frac{\partial S}{\partial x} = 0$ and $\frac{\partial S}{\partial y} = 0$), and solve the resulting simple linear simultaneous equations.

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