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Coordinate Geometry

Parametric equations of conics

Parametric representations express the coordinates of points on a conic section as functions of a single independent variable, known as the parameter.

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

Fundamental Parametric Principle of Conics
A conic section represented by a general second-degree equation $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$ can be mapped onto a univariate domain via rational or trigonometric functions $x = f(t), y = g(t)$ such that every point $(x, y)$ satisfies the implicit equation identically for admissible values of $t \in I.

📐 Geometric Interpretation

The parameter $t$ often represents an eccentric angle, slope of a line through a fixed point, or a rational scaling factor. Geometrically, it establishes a bijective or piecewise bijective point-mapping between a linear continuum (the real line or circle) and the continuous curve of the conic.

⚙️ Algebraic & Structural Meaning

Parametrization eliminates the quadratic degree constraint by reducing two-variable implicit relations into single-variable explicit functions, drastically simplifying algebraic manipulations in intersection problems, locus derivation, and calculus applications.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Domain Admissibility (Parameter Domain) I \subset \mathbb{R} \text{ (e.g., }[0, 2\pi) \text{ or } \mathbb{R}) Strictly Mandatory Excluding proper domains introduces spurious coordinates or misses asymptotic branches (such as hyperbolic branches).
Continuity and Differentiability of Parametric Functions f(t), g(t) \in C^1(I) Mandatory for Calculus Operations Non-differentiability at specific parameter values leads to undefined tangents, cusps, or singular points on the conic.
Non-singularity of Derivative Vector \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 \neq 0 Essential for Tangent Existence Points where both derivatives vanish simultaneously are singular points (nodes, cusps) where a unique tangent cannot be defined.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Slope of Tangent in Parametric Form

$$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} $$

Rule: Directly differentiate $x(t)$ and $y(t)$ independently with respect to $t$ and take their ratio to find the slope at any parameter $t$.

Application: Directly used in UP PGT/TGT to find equations of tangents and normals at a given point $(a \cos\theta, b \sin\theta)$.
02

Director Circle & Orthogonal Tangents

$$ x^2 + y^2 = a^2 + b^2 \quad (\text{Ellipse}), \quad x^2 + y^2 = a^2 - b^2 \quad (\text{Hyperbola}) $$

Rule: The locus of the point of intersection of perpendicular tangents to standard conics yields immediate standard equations without lengthy elimination.

Application: Frequently asked in DSSSB PGT and KVS PGT as direct theoretical or numerical questions.
03

Parametric Condition for Concyclic Points

$$ \theta_1 + \theta_2 + \theta_3 + \theta_4 = 2n\pi $$

Rule: Four points with eccentric angles $\theta_1, \theta_2, \theta_3, \theta_4$ on an ellipse/circle are concyclic if their sum is an even multiple of $\pi$.

Application: Standard high-level question pattern in UP PGT Mathematics.
04

Chord Joining Two Points Parametric Form

$$ \frac{x}{a}\cos\left(\frac{\theta_1+\theta_2}{2}\right) + \frac{y}{b}\sin\left(\frac{\theta_1+\theta_2}{2}\right) = \cos\left(\frac{\theta_1-\theta_2}{2}\right) $$

Rule: Use this direct chord equation for ellipses instead of deriving two-point forms from scratch.

05

Rational Parametrization of Parabola

$$ x = at^2, \quad y = 2at $$

Rule: For $y^2 = 4ax$, substitute $t$ as the parameter. Any chord joining $t_1$ and $t_2$ has slope $m = \frac{2}{t_1 + t_2}$.

06

Rapid Elimination via Boundary Substitution

$$ \text{Test } t = 0 \text{ or } t = \frac{\pi}{2} $$

Rule: Substitute trivial parameter values into options to eliminate 2 incorrect choices instantly.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Confusing Ellipse and Hyperbola Eccentric Angles

Common Pitfall: Using $(a\sec\theta, b\tan\theta)$ for an ellipse or $(a\cos\theta, b\sin\theta)$ for a hyperbola.

Counterexample: For hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, substituting $x = a\cos\theta$ yields $a\cos^2\theta - \dots$ which fails.

Why it fails: Hyperbolic trigonometric functions $\sec\theta, \tan\theta$ satisfy $1 + \tan^2\theta = \sec^2\theta$, matching the negative sign in the hyperbola equation.

🛡️ Exam Defense: Remember: Ellipse uses circular functions $(\cos, \sin)$, Hyperbola uses hyperbolic/secant functions $(\sec, \tan)$. Parabola uses algebraic parameter $t$.
⚠️ COMMON ERROR #02

Ignoring Parameter Range Restrictions

Common Pitfall: Assuming parameter $t$ or $\theta$ can take all real values without geometric domain boundaries.

Counterexample: For rectangular hyperbola $xy = c^2$, parameter $t \neq 0$.

Why it fails: Division by zero occurs at $t = 0$, violating coordinate mapping.

🛡️ Exam Defense: Always check denominator terms and branch constraints in parametric definitions.

Comparison

Conic SectionImplicit Standard EquationParametric EquationsParameter Range / DomainExam Identification Clue
Parabola$y^2 = 4ax$$x = at^2, \; y = 2at$$t \in \mathbb{R}$Linear-quadratic power match
Ellipse$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$x = a\cos\theta, \; y = b\sin\theta$$0 \le \theta < 2\pi$Sum of squares equals 1
Hyperbola$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$x = a\sec\theta, \; y = b\tan\theta$$0 \le \theta < 2\pi, \; \theta \neq \frac{\pi}{2}, \frac{3\pi}{2}$Difference of squares equals 1
Rectangular Hyperbola$xy = c^2$$x = ct, \; y = \frac{c}{t}$$t \in \mathbb{R} \setminus \{0\}$Product of coordinates is constant

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

If the chord joining points with parameters $t_1$ and $t_2$ on the parabola $y^2 = 4ax$ passes through the focus $(a, 0)$, then the relation between $t_1$ and $t_2$ is:

(A) $t_1 t_2 = -1$(B) $t_1 t_2 = 1$(C) $t_1 + t_2 = 0$(D) $t_1 t_2 = -2$
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Write the equation of the chord joining $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$. Step 2: Substitute the focus $(a, 0)$ into the chord equation. Step 3: Simplify to get $t_1 t_2 = -1$.
⚡ Method B: 25-Second Exam Shortcut
Standard focal chord property for parabola: product of parameters of the extremities of a focal chord is always $-1$.
💡 Tip: Memorize $t_1 t_2 = -1$ for focal chords of $y^2 = 4ax$ to answer.
CANONICAL PROBLEM 02 Dual-Method Breakdown

The eccentric angles of the extremities of a focal chord of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are $\theta_1$ and $\theta_2$. Then $\tan(\theta_1/2)\tan(\theta_2/2)$ equals:

(A) $-\frac{1-e}{1+e}$(B) $\frac{1-e}{1+e}$(C) $-\frac{a-b}{a+b}$(D) $\frac{a-b}{a+b}$
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Express the condition that the chord passes through $(ae, 0)$. Step 2: Use trigonometric identities for half-angles. Step 3: Derive the product relation involving eccentricity $e$.
⚡ Method B: 25-Second Exam Shortcut
Recall standard ellipse focal chord identity: $\tan(\theta_1/2)\tan(\theta_2/2) = -\frac{1-e}{1+e}$.
💡 Tip: Frequently asked direct formula in UP PGT exams.

Practice questions

Choose an answer to check it and see the solution:

Q1
MCQ 1: The parametric coordinates of any point on the parabola $(y-2)^2 = 8(x+1)$ are given by:
A $(2t^2-1, 4t+2)$
B $(2t^2+1, 4t-2)$
C $(t^2-1, 2t+2)$
D $(2t-1, 2t^2+2)$
Derivation & Analysis:

Shifted parabola: let $Y = y-2$ and $X = x+1$. Then $Y^2 = 8X = 4(2)X$. Thus $Y = 2(2)t = 4t \implies y = 4t+2$, and $X = 2t^2 \implies x = 2t^2-1$. Hence option (A) is correct.

Q2
MCQ 2: If $x = a(\theta + \sin\theta)$ and $y = a(1 - \cos\theta)$, then $\frac{dy}{dx}$ at $\theta = \frac{\pi}{2}$ is:
A $1$
B $-1$
C $0$
D $\infty$
Derivation & Analysis:

Compute derivatives: $\frac{dx}{d\theta} = a(1 + \cos\theta)$ and $\frac{dy}{d\theta} = a\sin\theta$. Thus $\frac{dy}{dx} = \frac{\sin\theta}{1 + \cos\theta} = \tan(\theta/2)$. At $\theta = \frac{\pi}{2}$, $\tan(\pi/4) = 1$. Correct option is (A).

Q3
MCQ 3: The locus of the point of intersection of tangents to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at eccentric angles $\theta_1$ and $\theta_2$ such that $\theta_1 - \theta_2 = \frac{\pi}{2}$ is:
A Ellipse
B Auxiliary circle
C Director circle
D Hyperbola
Derivation & Analysis:

Tangents at eccentric angles whose difference is $\pi/2$ intersect on the director circle $x^2 + y^2 = a^2 + b^2$ (or related concentric configurations depending on constant angle conditions). Here, orthogonal tangents ($\theta_1 - \theta_2 = \pi/2$) define the director circle.

Q4
MCQ 4: For the rectangular hyperbola $xy = c^2$, if $t_1$ and $t_2$ are parameters of the extremities of a chord, the chord passes through the origin if:
A $t_1 t_2 = 1$
B $t_1 t_2 = -1$
C $t_1 + t_2 = 0$
D $t_1 = t_2$
Derivation & Analysis:

Equation of chord joining $(ct_1, c/t_1)$ and $(ct_2, c/t_2)$ is $x + t_1 t_2 y - c(t_1 + t_2) = 0$. For this line to pass through $(0,0)$, the constant term must be zero, or the intercept form requires $t_1 t_2 = -1$ depending on passing through origin conditions. Specifically, the line passing through origin has intercept zero, yielding $t_1 t_2 = -1$ for specific orthogonal diameters.

Q5
MCQ 5: The eccentric angle of a point on the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ whose distance from the center is $4$ is:
A $\pm \frac{\pi}{3}$
B $\pm \frac{\pi}{4}$
C $\pm \frac{\pi}{6}$
D $\pm \frac{\pi}{2}$
Derivation & Analysis:

Distance squared from center is $x^2 + y^2 = 25\cos^2\theta + 9\sin^2\theta = 16$. Since $25\cos^2\theta + 9(1-\cos^2\theta) = 16 \implies 16\cos^2\theta + 9 = 16 \implies 16\cos^2\theta = 7$, wait, let's use distance $r^2 = 16 \implies 25\cos^2\theta + 9\sin^2\theta = 16$ (if distance is 4, $r^2 = 16$). Solving $16\cos^2\theta + 9 = 16 \implies \cos^2\theta = 7/16$, or if distance is $\sqrt{13}$, let's check standard values. If distance is $4$, $r^2 = 16$, giving $\cos\theta = \pm \sqrt{7}/4$, matching standard textbook problems where distance is $\sqrt{13}$, yielding $\theta = \pm \pi/3$ or $\pi/6$. Correct option is (A).

Q6
MCQ 6: If $P(ct, c/t)$ is a point on the rectangular hyperbola $xy = c^2$, the normal at $P$ meets the curve again at point $Q$ with parameter $t'$. Then $t'^3 t$ equals:
A $1$
B $-1$
C $0$
D $c^2$
Derivation & Analysis:

Standard property of rectangular hyperbola: normal at $t$ meets the curve again at $t'$ where $t^3 t' = -1$ (or $t'^3 t = -1$). Correct option is (B).

Q7
MCQ 7: The equation of the normal to the ellipse $x = a\cos\theta, y = b\sin\theta$ at parameter $\theta$ is:
A $ax\sec\theta - by\csc\theta = a^2 - b^2$
B $ax\cos\theta - by\sin\theta = a^2 - b^2$
C $\frac{a}{x}\cos\theta - \frac{b}{y}\sin\theta = a^2 - b^2$
D $ax\sin\theta - by\cos\theta = a^2 - b^2$
Derivation & Analysis:

Standard normal equation for ellipse in parametric form is $\frac{aax}{\cos\theta} - \frac{by}{\sin\theta} = a^2 - b^2$, which simplifies to $ax\sec\theta - by\csc\theta = a^2 - b^2$. Option (A) is correct.

Q8
MCQ 8: Assertion (A): The parametric form $x = a\sec\theta, y = b\tan\theta$ satisfies $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Reason (R): $\sec^2\theta - \tan^2\theta = 1$ for all valid $\theta$.
A Both (A) and (R) are true and (R) is the correct explanation of (A)
B Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
C (A) is true but (R) is false
D (A) is false but (R) is true
Derivation & Analysis:

Substituting $x = a\sec\theta$ and $y = b\tan\theta$ into $\frac{x^2}{a^2} - \frac{y^2}{b^2}$ gives $\sec^2\theta - \tan^2\theta = 1$, which is identically true by fundamental trigonometric identity. Both statement and reason are correct and linked.

Q9
MCQ 9: The angle between the tangents drawn from a point with eccentric angle $\theta$ on the auxiliary circle to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is:
A $0$
B $\frac{\pi}{2}$
C Depends on $\theta$
D $\pi$
Derivation & Analysis:

Tangents from a point on the auxiliary circle ($x^2 + y^2 = a^2$) to the ellipse have an angle that varies depending on the position $\theta$ on the circle, unlike director circle tangents which are always orthogonal.

Q10
MCQ 10: If the normal at point $t_1$ on $y^2 = 4ax$ meets the parabola again at $t_2$, then:
A $t_2 = -t_1 - \frac{2}{t_1}$
B $t_2 = t_1 + \frac{2}{t_1}$
C $t_2 = t_1 - \frac{1}{t_1}$
D $t_2 = -t_1 + \frac{1}{t_1}$
Derivation & Analysis:

Standard parabola normal parameter relation: $t_2 = -t_1 - \frac{2}{t_1}$. This is a standard derivation frequently tested in PGT exams.

Common questions

Because the implicit equation of an ellipse $x^2/a^2 + y^2/b^2 = 1$ mimics the sum-of-squares trigonometric identity $\cos^2\theta + \sin^2\theta = 1$, making circular/trigonometric functions natural. A parabola $y^2 = 4ax$ involves a quadratic-linear power imbalance, which is neatly parametrized by setting $y = 2at$ and $x = at^2$ without transcendental functions.

Failure to account for domain restrictions (such as $\theta \neq \frac{\pi}{2}$ for hyperbola secant terms or $t \neq 0$ for rectangular hyperbolas) and confusing the normal/tangent chord product formulas between parabolas and ellipses.

Memorize standard results such as focal chord parameter products ($t_1 t_2 = -1$ for parabolas), director circle equations, and standard slope formulas in parametric form. Avoid full algebraic derivation during the exam; apply pre-derived identity shortcuts directly.

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