4. Coordinate Geometry of Two Dimensions

Classification as hyperbola

The general equation of second degree in two variables represents a hyperbola if and only if the discriminant determinant $\Delta \neq 0$ and the invariant condition $h^2 > ab$ holds.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Discriminant Condition for Hyperbola

The general homogeneous and non-homogeneous second-degree equation given by $$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$$ represents a hyperbola if the determinant $$\Delta = \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} \neq 0$$ and the second-degree coefficient condition satisfies $h^2 > ab$. Furthermore, if $h^2 > ab$ and $a + b \neq 0$, it represents a non-rectangular hyperbola; if $h^2 > ab$ and $a + b = 0$, it represents a rectangular hyperbola.

📐 Geometric Interpretation

Geometrically, the condition $h^2 > ab$ ensures that the cutting plane intersects both nappes of a double right-circular cone, producing two unbounded, disjoint open branches opening in opposite directions. The asymptotes of the hyperbola have slopes given by the quadratic equation in slope form $bm^2 + 2hm + a = 0$.

⚙️ Algebraic & Structural Meaning

Algebriacally, the transformation of coordinates via rotation and translation eliminates the cross-product term $2hxy$, converting the general quadratic form into its canonical diagonalized principal-axis form where the squared terms have opposite signs.

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 ($\Delta \neq 0$)	`Global Determinant Condition`	Strictly Mandatory / Necessary	If $\Delta = 0$, the second-degree equation degenerates into a pair of intersecting straight lines, representing a degenerate conic rather than a proper hyperbola.
Discriminant Inequality ($h^2 > ab$)	`Coefficient Inequality`	Strictly Mandatory	If $h^2 = ab$, the conic becomes a parabola; if $h^2 < ab$, it becomes an ellipse (or circle/imaginary ellipse).
Real Coefficients ($a, b, h, g, f, c \in \mathbb{R}$)	`Real Domain`	Essential for real locus	If coefficients are complex or violate real locus constraints, the hyperbola may become imaginary (locus of no real points).

Useful results and shortcuts

Use these identities and results when their conditions apply:

Rapid Hyperbola Test via Determinant Expansion

$$\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 \neq 0 \quad \text{and} \quad h^2 > ab$$

Rule: Quickly evaluate $\Delta$ and $h^2 - ab$. If $\Delta \neq 0$ and $h^2 > ab$, instantly mark hyperbola.

Application: Frequently asked in UP PGT to identify conic sections from a general degree-2 equation.

Asymptote Slopes Formula

$$m_1, m_2 = \frac{-h \pm \sqrt{h^2 - ab}}{b} \quad (b \neq 0)$$

Rule: The slopes of the two asymptotes of the hyperbola are immediately found using the roots of $bm^2 + 2hm + a = 0$.

Application: Directly applied in DSSSB PGT and KVS PGT questions asking for angle between asymptotes or asymptote equations.

Rectangular Hyperbola Condition

$$a + b = 0 \quad \text{with} \quad h^2 > ab$$

Rule: If the sum of coefficients of $x^2$ and $y^2$ is zero ($a + b = 0$), the hyperbola is rectangular (asymptotes perpendicular).

Application: Standard recurring question type in UP TGT/PGT.

Center of the Hyperbola

$$x_0 = \frac{hf - bg}{ab - h^2}, \quad y_0 = \frac{gh - af}{ab - h^2}$$

Rule: Obtained by partially differentiating the general equation with respect to $x$ and $y$ and solving the linear system $\frac{\partial S}{\partial x} = 0$ and $\frac{\partial S}{\partial y} = 0$.

Eccentricity Formula from Coefficients

$$e = \sqrt{\frac{2\sqrt{(a-b)^2 + 4h^2}}{(a+b) \pm \sqrt{(a-b)^2 + 4h^2}}}$$

Rule: Direct relation to compute eccentricity directly from general equation parameters without shifting origin.

Application: Asked in high-level PGT screening tests.

Elimination via Degeneracy Check

$$\Delta = 0 \implies \text{Degenerate Conic (Lines)}$$

Rule: Always check $\Delta = 0$ first. If zero, eliminate all proper conic options (ellipse, parabola, hyperbola) instantly.

Application: Eliminates two options.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Forgetting the Determinant Non-Zero Condition

Common Pitfall: Candidates check only $h^2 > ab$ and conclude it is a hyperbola, ignoring whether $\Delta = 0$.

Counterexample: $xy = 0$

Why it fails: Here $a=0, b=0, h=1/2$, so $h^2 > ab$ ($1/4 > 0$), but $\Delta = 0$, meaning it represents a pair of intersecting lines (axes), not a hyperbola.

🛡️ Exam Defense: Always compute $\Delta$ alongside $h^2 - ab$ before concluding.

⚠️ COMMON ERROR #02

Mistaking Rectangular Hyperbola Definition

Common Pitfall: Assuming $xy = c^2$ is the only form of rectangular hyperbola and missing general rotated forms where $a+b=0$.

Counterexample: $x^2 - y^2 + 4xy - 2x + 4y - 5 = 0$

Why it fails: Here $a=1, b=-1$, so $a+b=0$. $h=2$, $h^2 > ab$ ($4 > -1$), and $\Delta \neq 0$. This is a rotated rectangular hyperbola.

🛡️ Exam Defense: Check $a+b=0$ and $h^2 > ab$ together.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Hyperbola	$h^2 > ab$ and $\Delta \neq 0$	$h^2 > ab$	Two unbounded open branches	Asymptotes real and distinct
Parabola	$h^2 = ab$ and $\Delta \neq 0$	$h^2 = ab$	Single open parabolic curve	Axis parallel to symmetry line
Ellipse	$h^2 < ab$ and $\Delta \neq 0$	$h^2 < ab$	Closed bounded oval curve	Real closed locus
Degenerate Conic	$\Delta = 0$	$\Delta = 0$	Intersecting or parallel lines	Determinant vanishes identically

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the condition for the equation $x^2 + 2hxy + y^2 + 2gx + 2fy + c = 0$ to represent a hyperbola.

(A) $h^2 > 1$(B) $h^2 < 1$(C) $h^2 = 1$(D) $h^2 \ge 1$

📚 Method A: Formal Textbook Derivation

Step 1: Identify coefficients $a=1, b=1, h=h$. Step 2: Apply hyperbola condition $h^2 > ab$. Step 3: Substitute $a=1, b=1$ to get $h^2 > (1)(1) = 1$. Step 4: Verify $\Delta \neq 0$.

⚡ Method B: 25-Second Exam Shortcut

Since $a=1$ and $b=1$, $ab=1$. Hence $h^2 > 1$ directly.

💡 Tip: Direct coefficient matching saves time.

CANONICAL PROBLEM 02 Dual-Method Breakdown

The equation $12x^2 - 10xy + 2y^2 + 11x - 5y + 2 = 0$ represents:

(A) A pair of straight lines(B) A parabola(C) An ellipse(D) A hyperbola

📚 Method A: Formal Textbook Derivation

Step 1: Extract $a=12, b=2, h=-5, g=11/2, f=-5/2, c=2$. Step 2: Compute $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2$. Step 3: Find $\Delta = 0$. Step 4: Conclude it is a degenerate conic (pair of lines).

⚡ Method B: 25-Second Exam Shortcut

Check $\Delta$ first. If $\Delta = 0$, it is degenerate (lines).

💡 Tip: Always test $\Delta=0$ for degree-2 equations with messy numbers.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: The general second-degree equation represents a hyperbola if:

A $h^2 = ab, \Delta \neq 0$

B $h^2 < ab, \Delta \neq 0$

C $h^2 > ab, \Delta \neq 0$

D $h^2 > ab, \Delta = 0$

Derivation & Analysis:

A hyperbola requires $h^2 > ab$ and non-zero discriminant determinant $\Delta \neq 0$.

MCQ 2: If the equation $ax^2 + 2hxy + by^2 = 0$ represents two real and distinct lines, then:

A $h^2 = ab$

B $h^2 > ab$

C $h^2 < ab$

D $h^2 \ge ab$

Derivation & Analysis:

For homogeneous second degree equation to represent real and distinct lines, $h^2 > ab$.

MCQ 3: The equation $x^2 - 4xy + y^2 = 0$ represents:

A An ellipse

B A parabola

C A pair of real lines forming a hyperbola family

D A rectangular hyperbola

Derivation & Analysis:

Here $a=1, b=1 \implies a+b=2 \neq 0$. Wait, for rectangular hyperbola $a+b=0$. Here $h=-2, ab=1$, $h^2 = 4 > 1$, and $a+b = 2$. This is a non-rectangular hyperbola.

MCQ 4: What is the value of $\Delta$ for a non-degenerate hyperbola?

A $\Delta = 0$

B $\Delta \neq 0$

C $\Delta < 0$

D $\Delta > 0$

Derivation & Analysis:

Non-degenerate conics always have $\Delta \neq 0$.

MCQ 5: The condition for $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ to be a rectangular hyperbola is:

A $h^2 > ab$ and $a+b=0$

B $h^2 = ab$ and $a+b=0$

C $h^2 < ab$ and $a+b=0$

D $h^2 > ab$ and $a=b$

Derivation & Analysis:

Rectangular hyperbola condition is $h^2 > ab$ and $a+b = 0$.

MCQ 6: If $h^2 < ab$ and $\Delta \neq 0$, the conic is:

A Parabola

B Hyperbola

C Ellipse

D Rectangular hyperbola

Derivation & Analysis:

When $h^2 < ab$, the conic is classified as an ellipse.

MCQ 7: The asymptotes of the general hyperbola $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ differ from the curve only by a constant:

A $c$

B $-c$

C $k$

D $0$

Derivation & Analysis:

The equation of asymptotes is given by $ax^2 + 2hxy + by^2 + 2gx + 2fy + c' = 0$ where $c'$ is determined using $\Delta' = 0$.

MCQ 8: Assertion (A): $x^2 - 5xy + 6y^2 = 0$ represents a pair of real distinct lines. Reason (R): $h^2 > ab$ for this equation.

A Both A and R are true and R is the correct explanation

B Both A and R are true but R is not the correct explanation

C A is true, R is false

D A is false, R is true

Derivation & Analysis:

Here $a=1, b=6, h=-5/2$. $h^2 = 25/4 = 6.25$ and $ab = 6$. Since $6.25 > 6$, $h^2 > ab$, representing real distinct lines.

MCQ 9: The eccentricity $e$ of a hyperbola satisfies:

A $e = 1$

B $0 < e < 1$

C $e > 1$

D $e = 0$

Derivation & Analysis:

By definition, for any hyperbola, eccentricity $e > 1$.

Q10

MCQ 10: If $h^2 = ab$ and $\Delta \neq 0$, the conic represents:

A Hyperbola

B Ellipse

C Parabola

D Circle

Derivation & Analysis:

The condition $h^2 = ab$ with $\Delta \neq 0$ defines a parabola.

Common questions

If $\Delta = 0$, the second-degree equation factors into two linear equations representing intersecting straight lines. Although $h^2 > ab$ holds for intersecting lines, they form a degenerate conic, not a proper hyperbola.

A rectangular hyperbola requires both $h^2 > ab$ and $a + b = 0$, which implies that its asymptotes intersect at right angles ($90^\circ$). A general hyperbola only requires $h^2 > ab$ without the orthogonal coefficient sum restriction.

First calculate $h^2 - ab$. If it is positive, check if the determinant $\Delta$ is non-zero. This eliminates parabolas and ellipses immediately and distinguishes proper hyperbolas from intersecting lines.

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