Coordinate Geometry

Conics using transformation of coordinate axes

This section covers transformation of coordinate axes via translation and rotation to simplify general second-degree equations into canonical conic.

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

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Definition and statement

Principal Axes Transformation Theorem for Conics

Any general second-degree equation in two variables, given by $$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$$, can be reduced to its canonical form $$AX^2 + BY^2 + K = 0$$ or $$AX^2 = 4LY$$ by a suitable translation of origin to the center $(h_0, k_0)$ and a subsequent rotation of axes through an angle $\theta$ given by $$\tan 2\theta = \frac{2h}{a - b}$$.

📐 Geometric Interpretation

Translation shifts the center of the conic to the new origin eliminating linear terms $x$ and $y$, while rotation aligns the coordinate axes with the principal axes of the conic, eliminating the cross-product term $xy$.

⚙️ Algebraic & Structural Meaning

Under orthogonal transformations (translation and rotation), the discriminant $h^2 - ab$, the trace $a + b$, and the determinant $\Delta = \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}$ remain invariant, ensuring the nature of the conic is preserved.

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 Discriminant for Central Conics	$ab - h^2 \neq 0$	Strictly Mandatory	If $ab - h^2 = 0$, the conic is a parabola (or degenerate parabola/parallel lines), and a unique center $(h_0, k_0)$ does not exist; translation alone cannot eliminate linear terms.
Orthogonality of Rotation	$\theta \in [0, \pi/2)$	Strictly Mandatory	Failure to choose the correct quadrant for $\theta$ using $\tan 2\theta = \frac{2h}{a-b}$ results in axes being swapped (semi-major and semi-minor axes interchanged).
Non-Degeneracy Condition	$\Delta \neq 0$	Essential for proper conics	If $\Delta = 0$, the equation represents a degenerate conic (pair of intersecting lines, coincident lines, or a point).

Useful results and shortcuts

Use these identities and results when their conditions apply:

Elimination of Cross-Product Term Angle

$$\theta = \frac{1}{2}\tan^{-1}\left(\frac{2h}{a - b}\right)$$

Rule: If $a = b$, immediately set $\theta = \frac{\pi}{4}$ ($45^\circ$) without calculating inverse trigonometric functions.

Application: Frequently asked in UP PGT to quickly identify the rotation angle for equations like $x^2 + 4xy + y^2 = 1$.

Invariance of Trace and Discriminant

$$A + B = a + b, \quad AB - H^2 = ab - h^2$$

Rule: Use the sum of roots of the characteristic equation $\lambda^2 - (a+b)\lambda + (ab-h^2) = 0$ to instantly find the coefficients $A$ and $B$ of the transformed equation $AX^2 + BY^2 + K = 0$.

Finding the Center of a Conic

$$\frac{\partial F}{\partial x} = 0, \quad \frac{\partial F}{\partial y} = 0$$

Rule: Partial differentiate the general quadratic expression with respect to $x$ and $y$ and solve the linear simultaneous equations to find the center $(h_0, k_0)$ instantly.

Application: Directly tested in UP TGT for finding shifted origins.

Constant Term Determination After Translation

$$K = gx_0 + fy_0 + c$$

Rule: The constant term $K$ in the canonical equation equals half the linear terms evaluated at the center plus the original constant $c$.

Application: Crucial for writing the exact equation of ellipse/hyperbola after shifting origin.

Nature of Conic via Discriminant

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

Rule: Never transform axes to check the type of conic; inspect $h^2 - ab$ directly from the general equation.

Rapid Option Elimination via Determinant $\Delta$

$$\Delta = abc + 2fgh - af^2 - bg^2 - ch^2$$

Rule: Calculate $\Delta$ first. If $\Delta = 0$, eliminate all options representing non-degenerate ellipses or hyperbolas.

Application: Eliminates 2 out of 4 options instantly in UP PGT coordinate geometry questions.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Confusing Rotation Angle Quadrant

Common Pitfall: Using $\tan 2\theta = \frac{2h}{a-b}$ and picking an angle $\theta$ that does not satisfy $H = 0$ after transformation.

Counterexample: For $a < b$, if $\tan 2\theta$ is positive, $2\theta$ lies in the first quadrant, so $\theta \in (0, \pi/4)$.

Why it fails: Taking the wrong branch of inverse tangent flips the major and minor axes, leading to inverted focal properties.

🛡️ Exam Defense: Always check that the coefficient of $XY$ vanishes: $H = (b-a)\sin\theta\cos\theta + h(\cos^2\theta - \sin^2\theta) = 0$.

⚠️ COMMON ERROR #02

Ignoring Degeneracy when $\Delta = 0$

Common Pitfall: Treating an equation with $\Delta = 0$ as a standard ellipse or hyperbola after completing squares.

Counterexample: $x^2 + 2xy + y^2 - 2x - 2y + 1 = 0$, where $\Delta = 0$.

Why it fails: This factors into $(x+y-1)^2 = 0$, which represents coincident parallel lines, not a true conic.

🛡️ Exam Defense: Always evaluate the determinant $\Delta$ before attempting axis transformation.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Translation of Axes	New axes parallel to original	$x = X + h, y = Y + k$	Shifting origin without changing orientation	Elimination of $x$ and $y$ terms
Rotation of Axes	Origin fixed, axes rotated by $\theta$	$x = X\cos\theta - Y\sin\theta$	Turning axes to align with principal directions	Elimination of $xy$ term
General Conic Reduction	$ab - h^2 \neq 0, \Delta \neq 0$	$AX^2 + BY^2 + K = 0$	Standardizing ellipse/hyperbola canonical form	Mixed $xy$ and linear terms present
Parabola Reduction	$ab - h^2 = 0, \Delta \neq 0$	$AX^2 = 4LY$ or $BY^2 = 4LX$	Shifting vertex to origin and aligning axis	Parabolic second-degree terms with linear remainder

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the angle through which the coordinate axes must be rotated to remove the $xy$ term from the equation $x^2 - 2\sqrt{3}xy + y^2 = 2a^2$.

(A) $\frac{\pi}{6}$(B) $\frac{\pi}{4}$(C) $\frac{\pi}{3}$(D) $\frac{\pi}{12}$

📚 Method A: Formal Textbook Derivation

Here $a = 1, b = 1, 2h = -2\sqrt{3} \implies h = -~\sqrt{3}$. Using $\tan 2\theta = \frac{2h}{a-b}$, since $a - b = 0$, $\tan 2\theta = \infty \implies 2\theta = \frac{\pi}{2} \implies \theta = \frac{\pi}{4}$. Wait, re-evaluating: $\tan 2\theta = \frac{-2\sqrt{3}}{1-1} = \infty \implies 2\theta = \frac{\pi}{2} \implies \theta = \frac{\pi}{4}$. Let us check: if $a=b$, $\theta = \pi/4$. Correct option is (B).

⚡ Method B: 25-Second Exam Shortcut

Standard rule: When coefficient of $x^2$ equals coefficient of $y^2$ ($a=b$), the required rotation angle is always $\theta = \pi/4$.

💡 Tip: Do not waste time calculating inverse trig when $a=b$; mark $\pi/4$ instantly.

CANONICAL PROBLEM 02 Dual-Method Breakdown

If the origin is shifted to $(1, 2)$ without rotation, what does the equation $x^2 + y^2 - 2x - 4y + 1 = 0$ transform to?

(A) $X^2 + Y^2 = 4$(B) $X^2 + Y^2 = 3$(C) $X^2 + Y^2 = 1$(D) $X^2 + Y^2 = 5$

📚 Method A: Formal Textbook Derivation

Substitute $x = X + 1$ and $y = Y + 2$ into the equation: $(X+1)^2 + (Y+2)^2 - 2(X+1) - 4(Y+2) + 1 = 0$. Expanding yields $X^2 + 2X + 1 + Y^2 + 4Y + 4 - 2X - 2 - 4Y - 8 + 1 = 0 \implies X^2 + Y^2 + (1 + 4 - 2 - 8 + 1) = 0 \implies X^2 + Y^2 - 4 = 0 \implies X^2 + Y^2 = 4$.

⚡ Method B: 25-Second Exam Shortcut

Recognize that the given equation represents a circle with center $(1,2)$ and radius $r = \sqrt{1^2 + 2^2 - 1} = 2$. Shifting the origin to the center directly gives $X^2 + Y^2 = r^2 = 4$.

💡 Tip: Always check if the given equation represents a standard circle or conic; shifting origin to the center eliminates linear terms directly.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: The invariant quantity under both translation and rotation of coordinate axes for the general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is:

A $a + b$

B $ab - h^2$

C $h^2 - 4ab$

D Both (A) and (B)

Derivation & Analysis:

Both the sum of coefficients $a+b$ (trace) and the discriminant $ab-h^2$ are fundamental invariants under orthogonal transformations of coordinate axes.

MCQ 2: To eliminate the first-degree terms $x$ and $y$ from the equation $2x^2 + 3y^2 - 4x + 12y - 1 = 0$, the origin must be shifted to:

A $(1, -2)$

B $(-1, 2)$

C $(1, 2)$

D $(-1, -2)$

Derivation & Analysis:

Differentiating partially with respect to $x$: $4x - 4 = 0 \implies x = 1$. Differentiating with respect to $y$: $6y + 12 = 0 \implies y = -2$. Thus, origin shifts to $(1, -2)$.

MCQ 3: If the axes are rotated through an angle of $30^\circ$, the transformed coordinates of $(1, \sqrt{3})$ are:

A $(2, 0)$

B $(0, 2)$

C $(1, 1)$

D $(\sqrt{3}, 1)$

Derivation & Analysis:

Using $X = x\cos\theta + y\sin\theta$ and $Y = -x\sin\theta + y\cos\theta$ with $\theta = 30^\circ$: $X = 1(\frac{\sqrt{3}}{2}) + \sqrt{3}(\frac{1}{2}) = \sqrt{3}$, and $Y = -1(\frac{1}{2}) + \sqrt{3}(\frac{\sqrt{3}}{2}) = 1$. Wait: $X = 1(\frac{\sqrt{3}}{2}) + \sqrt{3}(\frac{1}{2}) = \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \sqrt{3}$. Let's re-verify: $x=1, y=\sqrt{3}$, $X = x\cos 30 + y\sin 30 = 1(\frac{\sqrt{3}}{2}) + \sqrt{3}(\frac{1}{2}) = \sqrt{3}$. $Y = -1(\frac{1}{2}) + \sqrt{3}(\frac{\sqrt{3}}{2}) = -\frac{1}{2} + \frac{3}{2} = 1$. So $(\sqrt{3}, 1)$. Let's check options: Option (D).

MCQ 4: The equation $xy = c^2$ when rotated through an angle of $45^\circ$ transforms into:

A $X^2 - Y^2 = 2c^2$

B $X^2 + Y^2 = 2c^2$

C $\frac{X^2}{c^2} + \frac{Y^2}{c^2} = 1$

D $X^2 - Y^2 = c^2$

Derivation & Analysis:

Substituting $x = \frac{X - Y}{\sqrt{2}}$ and $y = \frac{X + Y}{\sqrt{2}}$ into $xy = c^2$ gives $\left(\frac{X-Y}{\sqrt{2}}\right)\left(\frac{X+Y}{\sqrt{2}}\right) = c^2 \implies \frac{X^2 - Y^2}{2} = c^2 \implies X^2 - Y^2 = 2c^2$.

MCQ 5: What is the nature of the conic represented by $3x^2 + 10xy + 3y^2 - 2x - 4y + 1 = 0$?

A Ellipse

B Parabola

C Hyperbola

D Rectangular Hyperbola

Derivation & Analysis:

Here $a=3, b=3, h=5$. Discriminant $h^2 - ab = 5^2 - (3)(3) = 25 - 9 = 16 > 0$. Since $h^2 - ab > 0$ and $a+b = 6 \neq 0$, it represents a hyperbola.

MCQ 6: If the coordinate axes are rotated by $45^\circ$, what is the new equation of $x^2 - y^2 = a^2$?

A $XY = \frac{a^2}{2}$

B $XY = a^2$

C $X^2 + Y^2 = a^2$

D $X^2 - Y^2 = 2a^2$

Derivation & Analysis:

Substituting $x = \frac{X - Y}{\sqrt{2}}, y = \frac{X + Y}{\sqrt{2}}$ into $x^2 - y^2 = a^2$ yields $\left(\frac{X-Y}{\sqrt{2}}\right)^2 - \left(\frac{X+Y}{\sqrt{2}}\right)^2 = a^2 \implies \frac{(X^2 - 2XY + Y^2) - (X^2 + 2XY + Y^2)}{2} = a^2 \implies -4XY = 2a^2 \implies XY = -\frac{a^2}{2}$, or $XY = \frac{a^2}{2}$ depending on rotation direction convention.

MCQ 7: The condition that the general second-degree equation represents a pair of straight lines is:

A $\Delta = 0$ and $h^2 \ge ab$

B $\Delta \neq 0$

C $h^2 = ab$ and $\Delta = 0$

D $\Delta = 0$ and $h^2 = ab$

Derivation & Analysis:

A general second-degree equation represents a degenerate conic (pair of straight lines) if and only if the determinant $\Delta = 0$, provided $h^2 \ge ab$ for real lines.

MCQ 8: If the origin is shifted to $(2, -3)$ and axes are rotated by $90^\circ$, the transformation equations for $(x, y)$ in terms of $(X, Y)$ are:

A $x = -Y + 2, y = X - 3$

B $x = X + 2, y = Y - 3$

C $x = Y + 2, y = -X - 3$

D $x = -X + 2, y = -Y - 3$

Derivation & Analysis:

For rotation by $90^\circ$, $x' = X\cos 90 - Y\sin 90 = -Y$ and $y' = X\sin 90 + Y\cos 90 = X$. With translation by $(2, -3)$, $x = -Y + 2$ and $y = X - 3$.

MCQ 9: The sum of the squares of the semi-axes of the conic $ax^2 + 2hxy + by^2 = 1$ is given by:

A $\frac{a+b}{ab-h^2}$

B $\frac{a-b}{ab-h^2}$

C $\frac{1}{a+b}$

D $\frac{a+b}{(ab-h^2)^2}$

Derivation & Analysis:

The sum of the squares of reciprocals of semi-axes equals $a+b$, and their product equals $ab-h^2$. Sum of semi-axes squared involves trace and determinant invariants.

Q10

MCQ 10: In transforming axes, if the transformed equation has no $XY$ term, the angle of rotation $\theta$ satisfies:

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

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

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

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

Derivation & Analysis:

The fundamental angle formula to eliminate the cross product term $xy$ is $\tan 2\theta = \frac{2h}{a-b}$.

Common questions

Translation eliminates the linear terms ($x$ and $y$) by placing the origin at the center of the conic. Rotation eliminates the cross-product term ($xy$), aligning the coordinate axes with the principal axes of symmetry (major and minor axes).

Candidates often miscalculate $\theta$ when $a = b$. When $a = b$, $\tan 2\theta = \infty$, which immediately dictates $\theta = \pi/4$ ($45^\circ$), rather than attempting to apply inverse trigonometric formulas incorrectly.

Yes! Always examine the discriminant $h^2 - ab$ and the determinant $\Delta$. This bypasses lengthy algebraic transformations and saves precious time in competitive exams.

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