Coordinate Geometry

Parametric equation of tangent

The parametric equation of a tangent line expresses the coordinates of any point on the tangent in terms of a single parameter, derived directly from the differential properties of conics.

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

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Definition and statement

Parametric Form of Tangent Equation

If a conic is represented parametrically by $x = f(t)$ and $y = g(t)$, the equation of the tangent at point $t$ is given by $Y - g(t) = \frac{g'(t)}{f'(t)} (X - f(t))$, provided $f'(t) \neq 0$.

📐 Geometric Interpretation

The parameter $t$ traces the conic curve continuously. At a fixed value of $t$, the derivative $\frac{dy}{dx} = \frac{g'(t)}{f'(t)}$ defines the instantaneous slope of the tangent line touching the curve at the parametric point $(f(t), g(t))$.

⚙️ Algebraic & Structural Meaning

Translates non-linear implicit or explicit algebraic curves into linear equations parameterized by a single variable, drastically simplifying intersection problems, chord lengths, and locus derivations.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Parametric Differentiability	`(t_1, t_2) (Open parameter interval)`	Strictly Mandatory / Necessary	If $f'(t)$ or $g'(t)$ fails to exist, the slope becomes indeterminate or infinite, failing to yield a unique tangent.
Non-zero Derivative Condition	`f'(t) \neq 0`	Strictly Mandatory	If $f'(t) = 0$ while $g'(t) \neq 0$, the tangent becomes vertical with equation $X = f(t)$. Division by zero occurs if standard slope form is unadjusted.
Curve Smoothness	`[f(t), g(t)] \in \mathbb{R}^2`	Essential for geometric validity	Cusps or self-intersections where $f'(t) = g'(t) = 0$ lead to multiple tangents or no well-defined tangent.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Parabola $y^2 = 4ax$ Parametric Tangent

$$ty = x + at^2$$

Rule: Substitute $x = at^2, y = 2at$ into the standard tangent formula $yy_1 = 2a(x + x_1)$ instantly yielding $ty = x + at^2$.

Application: undefined

Ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ Parametric Tangent

$$\frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1$$

Rule: Replace $x^2$ with $x x_1$ and use parametric coordinates $x_1 = a \cos \theta, y_1 = b \sin \theta$.

Application: Frequently used in TGT/PGT for finding length of intercept between axes.

Hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ Parametric Tangent

$$\frac{x \sec \phi}{a} - \frac{y \tan \phi}{b} = 1$$

Rule: Utilize hyperbolic parameters $x = a \sec \phi, y = b \tan \phi$ to write the tangent by T-formula substitution.

Application: Appears in questions calculating the area of the triangle formed by the tangent with coordinate axes.

Rectangular Hyperbola $xy = c^2$ Parametric Tangent

$$xt + yt = 2c \implies x + yt^2 = 2ct$$

Rule: Parametric coordinates $(ct, \frac{c}{t})$ give the tangent equation $x + yt^2 = 2ct$.

Application: High frequency in DSSSB PGT for normal and tangent intersection properties.

Point of Intersection of Tangents at $t_1$ and $t_2$ on Parabola

$$x = a t_1 t_2, \quad y = a(t_1 + t_2)$$

Rule: Solve the two tangent equations $ty = x + at^2$ simultaneously to find the intersection point without full derivation.

Application: Standard shortcut for locus and chord problems in UP PGT.

Condition of Tangency for Line $y = mx + c$ on Parabola

$$c = \frac{a}{m}$$

Rule: Equate the discriminant of the quadratic resulting from line-curve substitution to zero, or use slope $m = \frac{1}{t}$.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Confusing Parabola Parameters with Ellipse Angles

Common Pitfall: Applying trigonometric parameters $(\cos \theta, \sin \theta)$ to parabolas or algebraic parameters $t$ to ellipses.

Counterexample: Using $x \cos \theta + y \sin \theta = a$ for $y^2 = 4ax$

Why it fails: Parabolas are unbounded and use algebraic parameter $t \in \mathbb{R}$, whereas ellipses are bounded and use angular parameter $\theta \in [0, 2\pi)$.

🛡️ Exam Defense: Memorize curve-parameter pairings: Parabola $\to t$, Ellipse $\to \theta$, Hyperbola $\to \phi$, Rectangular Hyperbola $\to t$.

⚠️ COMMON ERROR #02

Ignoring Vertical Tangent Conditions

Common Pitfall: Assuming $\frac{dy}{dx}$ is always finite when finding parametric tangents.

Counterexample: At $t = 0$ for $x = t^3, y = t^2$

Why it fails: Here $f'(t) = 3t^2$, which vanishes at $t=0$, resulting in a vertical tangent ($x = 0$) where slope form $\frac{g'(t)}{f'(t)}$ blows up.

🛡️ Exam Defense: Always check both $f'(t) = 0$ and $g'(t) = 0$ simultaneously before writing slope-intercept equations.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Parabola Tangent	y^2 = 4ax, point (at^2, 2at)	ty = x + at^2	Line touching parabola at single point	Keywords: parabola, parameter t
Ellipse Tangent	x^2/a^2 + y^2/b^2 = 1, point (a \cos\theta, b \sin\theta)	(x \cos\theta)/a + (y \sin\theta)/b = 1	Line touching ellipse within bounded box	Keywords: eccentric angle, ellipse
Hyperbola Tangent	x^2/a^2 - y^2/b^2 = 1, point (a \sec\phi, b \tan\phi)	(x \sec\phi)/a - (y \tan\phi)/b = 1	Line touching hyperbola branch	Keywords: hyperbola, secant parameter
Rectangular Hyperbola	xy = c^2, point (ct, c/t)	x + yt^2 = 2ct	Tangent to rectangular hyperbola	Keywords: xy = constant, product of coordinates

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the equation of the tangent to the parabola $y^2 = 12x$ at the point corresponding to parameter $t = 2$.

(A) $2x - y + 6 = 0$(B) $x - 2y + 12 = 0$(C) $2x - y - 6 = 0$(D) $x - 2y - 12 = 0$

📚 Method A: Formal Textbook Derivation

Here $4a = 12 \implies a = 3$. The parametric coordinates are $(at^2, 2at) = (3(2^2), 2(3)(2)) = (12, 12)$. Using tangent formula $ty = x + at^2$, we get $2y = x + 3(2^2) \implies 2y = x + 12 \implies x - 2y + 12 = 0$.

⚡ Method B: 25-Second Exam Shortcut

Substitute $a = 3, t = 2$ directly into $ty = x + at^2 \implies 2y = x + 3(4) \implies x - 2y + 12 = 0$. Takes 15 seconds.

💡 Tip: Check option signs carefully; standard form is $x - ty + at^2 = 0$.

CANONICAL PROBLEM 02 Dual-Method Breakdown

If the tangents drawn at points $t_1$ and $t_2$ on the parabola $y^2 = 4ax$ intersect on the axis of the parabola, then:

(A) $t_1 + t_2 = 0$(B) $t_1 t_2 = 1$(C) $t_1 = t_2$(D) $t_1 t_2 = -1$

📚 Method A: Formal Textbook Derivation

The intersection point of tangents at $t_1$ and $t_2$ is $(at_1 t_2, a(t_1 + t_2))$. If it lies on the axis of the parabola ($y = 0$), then the ordinate must be zero. Hence $a(t_1 + t_2) = 0 \implies t_1 + t_2 = 0$.

⚡ Method B: 25-Second Exam Shortcut

Intersection ordinate is $y = a(t_1 + t_2)$. Axis of parabola means $y = 0 \implies t_1 + t_2 = 0$.

💡 Tip: A direct standard result frequently asked without modification in UP PGT.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: The parametric equation of the tangent to the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ at $\theta = \frac{\pi}{4}$ is:

A $\frac{x}{4} + \frac{y}{3} = \frac{1}{\sqrt{2}}$

B $\frac{x \sqrt{2}}{4} + \frac{y \sqrt{2}}{3} = 1$

C $3x + 4y = 12\sqrt{2}$

D Both (B) and (C)

Derivation & Analysis:

Substitute $\theta = \pi/4, a = 4, b = 3$ into $\frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1$. Since $\cos(\pi/4) = \sin(\pi/4) = 1/\sqrt{2}$, we get $\frac{x}{4\sqrt{2}} + \frac{y}{3\sqrt{2}} = 1 \implies \frac{x \sqrt{2}}{4} + \frac{y \sqrt{2}}{3} = 1$, which simplifies to $3x + 4y = 12\sqrt{2}$.

MCQ 2: If the tangent at point $t$ on $y^2 = 4ax$ meets the axis of the parabola at $T$, then the triangle formed by $T$, the vertex, and the point $t$ has area:

A $a^2 t^3$

B $\frac{1}{2} a^2 t^3$

C $2 a^2 t^3$

D $a^2 t^2$

Derivation & Analysis:

Tangent $ty = x + at^2$ meets axis ($y=0$) at $T(-at^2, 0)$. Vertex is $(0,0)$ and point on parabola is $(at^2, 2at)$. Area of triangle = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} (at^2) (2at) = a^2 t^3$. Wait, base on x-axis is from $-at^2$ to $at^2$ if measured from vertex? Let's check coordinates: Vertex $O(0,0)$, $T(-at^2,0)$, $P(at^2,2at)$. Base $OT = at^2$ (distance from origin to $T$ is $at^2$), height = $2at$. Area = $\frac{1}{2}(at^2)(2at) = a^2 t^3$. Correct option (A).

MCQ 3: The locus of the point of intersection of perpendicular tangents to the parabola $y^2 = 4ax$ is:

A $x = a$

B $x + a = 0$

C $y = 0$

D $x = 0$

Derivation & Analysis:

Tangents at $t_1$ and $t_2$ intersect at $(at_1 t_2, a(t_1 + t_2))$. For perpendicular tangents, $t_1 t_2 = -1$. Thus the x-coordinate is $x = a(-1) = -a$, which is the directrix $x + a = 0$.

MCQ 4: For the rectangular hyperbola $xy = c^2$, the equation of the tangent at parameter $t$ is:

A $x + yt^2 = 2ct$

B $xt + y = 2ct$

C $x - yt^2 = 2ct$

D $tx + y = c$

Derivation & Analysis:

Differentiating $xy = c^2$, $\frac{dy}{dx} = -\frac{y}{x}$. At parametric point $(ct, c/t)$, slope is $-1/t^2$. Using point-slope form: $y - \frac{c}{t} = -\frac{1}{t^2}(x - ct) \implies t^2 y - ct = -x + ct \implies x + t^2 y = 2ct$.

MCQ 5: The product of the perpendiculars from the foci upon any tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is:

A $a^2$

B $b^2$

C $ab$

D $e^2$

Derivation & Analysis:

A standard property of ellipses states that the product of perpendicular distances from the foci to any tangent is constant and equals $b^2$.

MCQ 6: If the line $lx + my + n = 0$ touches the parabola $y^2 = 4ax$, then:

A $am^2 = ln$

B $al^2 = mn$

C $an^2 = lm$

D $ln = am^2$

Derivation & Analysis:

Comparing $y = -\frac{l}{m}x - \frac{n}{m}$ with standard tangent $y = mx + c$ where $c = \frac{a}{m'}$, here slope $m' = -l/m$ and intercept $= -n/m$. Thus $-\frac{n}{m} = \frac{a}{-l/m} = \frac{am}{l} \implies -ln = am^2 \implies am^2 = ln$.

MCQ 7: The eccentric angles of the extremities of latus rectum of an ellipse are given by:

A $\tan \theta = \pm \frac{b}{ae}$

B $\tan \theta = \pm e$

C $\tan \theta = \pm \frac{ae}{b}$

D $\cos \theta = \pm e$

Derivation & Analysis:

The coordinates of the focus are $(ae, 0)$. Latus rectum ordinate is $b^2/a$. Solving for parametric point $(a \cos \theta, b \sin \theta)$ where $y = b^2/a$ yields $\tan \theta = \pm \frac{ae}{b}$.

MCQ 8: Assertion (A): The tangent at point $t_1$ and $t_2$ on $y^2 = 4ax$ intersect at $x = at_1 t_2, y = a(t_1 + t_2)$. Reason (R): Solving two linear parametric tangent equations simultaneously yields this coordinate.

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

C (A) is true but (R) is false

D (A) is false but (R) is true

Derivation & Analysis:

Both statements are mathematically rigorous and standard results in conic sections.

MCQ 9: The parameter $t$ for the point on the parabola $y^2 = 8x$ where the tangent is inclined at $45^circ$ to the axis of x is:

A $1$

B $2$

C $\frac{1}{2}$

D $4$

Derivation & Analysis:

Slope $m = \tan(45^circ) = 1$. For $y^2 = 4ax$ ($4a=8 \implies a=2$), the slope of the parametric tangent is $\frac{1}{t}$. Thus $\frac{1}{t} = 1 \implies t = 1$.

Q10

MCQ 10: The chord joining points $t_1$ and $t_2$ on the parabola $y^2 = 4ax$ passes through the focus if:

A $t_1 t_2 = 1$

B $t_1 t_2 = -1$

C $t_1 + t_2 = 0$

D $t_1 t_2 = 2$

Derivation & Analysis:

A focal chord of a parabola connecting parametric points $t_1$ and $t_2$ always satisfies the condition $t_1 t_2 = -1$.

Common questions

Parabolas are unbounded open curves extending to infinity, making an algebraic parameter $t \in \mathbb{R}$ ideal for tracking points. Ellipses are closed, bounded curves naturally parametrized by circular/trigonometric functions $(\cos \theta, \sin \theta)$ tied to their auxiliary circles.

Sign errors and confusing the parameter formulas between parabolas ($ty = x + at^2$) and hyperbolas. Candidates often misapply trigonometric identities when algebraic parameters are required.

Memorize standard intersection formulas (e.g., intersection of tangents at $t_1, t_2$ on a parabola is $(at_1t_2, a(t_1+t_2))$) to eliminate intermediate algebraic steps entirely.

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