Coordinate Geometry

Condition for tangent in limiting position

The limiting position of a chord or secant of a conic yields the equation of the tangent as two intersecting points coalesce into a single point.

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

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Definition and statement

Principle of Limiting Secant to Tangent

Let a curve be given by $f(x, y) = 0$. Consider two distinct points $P(x_1, y_1)$ and $Q(x_2, y_2)$ on the curve. The chord $PQ$ has the equation $$y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$$. Taking the limit as $Q \to P$ along the curve, the chord $PQ$ becomes the tangent at $P$, and the slope $\frac{y_2 - y_1}{x_2 - x_1}$ approaches the derivative $\frac{dy}{dx}\Big|_{(x_1, y_1)}$.

📐 Geometric Interpretation

Geometrically, as the secant line rotates about point $P$ while point $Q$ slides along the curve toward $P$, the secant assumes a unique limiting position called the tangent line at $P$, representing the instantaneous direction of the curve.

⚙️ Algebraic & Structural Meaning

Algebrically, the condition for a line to be a tangent is obtained by substituting the line equation into the conic equation, yielding a quadratic in one variable, and setting its discriminant to zero (or requiring coincident roots).

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Continuity of the curve in a neighborhood of $P$	$[x_1 - \delta, x_1 + \delta]$	Strictly Mandatory	If the curve has a break or jump discontinuity at $P$, the limiting position of the secant line is not unique or fails to exist.
Differentiability at the point of tangency $P$	`Open neighborhood containing $P$`	Strictly Mandatory	A sharp corner or cusp (such as $y = \|x\|$) results in distinct left-hand and right-hand limiting slopes, preventing a unique tangent.
Non-zero derivative or finite slope condition	`Point $P(x_1, y_1)$`	Essential for explicit slope form	If $\frac{dy}{dx} \to \infty$, the tangent becomes vertical ($x = x_1$), requiring the inverse reciprocal slope formulation $\frac{dx}{dy} = 0$.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Condition of Tangency for General Conic

$$abc + 2fgh - af^2 - bg^2 - ch^2 = 0$$

Rule: For the general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ to represent a pair of straight lines (degenerate conic), or for a line $lx + my + n = 0$ to touch the conic, use the determinant of the $3 \times 3$ coefficient matrix equal to zero.

Application: Directly tested in UP PGT to find parameters when a line touches a general conic.

Tangent to Conic $S = 0$ from Point $(x_1, y_1)$

$$T = 0$$

Rule: Replace $x^2$ by $xx_1$, $y^2$ by $yy_1$, $2xy$ by $xy_1 + yx_1$, $2x$ by $x + x_1$, $2y$ by $y + y_1$ in the equation of the conic to get the chord of contact or tangent equation instantly.

Application: Saves 2 minutes in TGT coordinate geometry calculations.

Condition for Line $y = mx + c$ to touch Parabola $y^2 = 4ax$

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

Rule: Substitute $y = mx + c$ into $y^2 = 4ax$, set discriminant $\Delta = 0$ of the resulting quadratic in $x$, yielding the exact tangency condition instantly.

Application: Frequent 1-mark question in DSSSB and KVS PGT Mathematics.

Condition for Line $y = mx + c$ to touch Ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

$$c^2 = a^2 m^2 + b^2$$

Rule: Directly memorize and apply the ellipse tangency condition without solving simultaneous quadratic equations.

Application: Standard recurring question pattern in UP PGT.

Condition for Line $y = mx + c$ to touch Hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

$$c^2 = a^2 m^2 - b^2$$

Rule: Note the sign change from the ellipse formula ($+b^2$ to $-b^2$) for hyperbolas.

Application: Frequently paired with ellipse questions to test sign confusion.

Point of Contact Coordinates for Parabola

$$x_1 = \frac{a}{m^2}, \quad y_1 = \frac{2a}{m}$$

Rule: Instantly find the exact point of contact when the tangent slope $m$ to $y^2 = 4ax$ is given.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Discriminant Zero Condition on Non-Quadratic Equations

Common Pitfall: Applying $\Delta = 0$ blindly to higher-degree algebraic curves or transcendental functions where the resulting substitution does not yield a simple quadratic.

Counterexample: $y = e^x$ and $y = mx + c$

Why it fails: Setting discriminant to zero only applies when the equation reduces to a strict quadratic in one variable.

🛡️ Exam Defense: Use calculus ($\frac{dy}{dx} = m$ and curve satisfaction) for transcendental or higher-degree curves.

⚠️ COMMON ERROR #02

Confusing Ellipse and Hyperbola Tangency Constants

Common Pitfall: Using $c^2 = a^2 m^2 + b^2$ for hyperbolas instead of $c^2 = a^2 m^2 - b^2$.

Counterexample: $c^2 = a^2 m^2 + b^2$ applied to $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

Why it fails: Leads to incorrect negative sign under square roots or wrong option selection.

🛡️ Exam Defense: Memorize the hyperbola condition with the negative sign: $c^2 = a^2 m^2 - b^2$.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Parabola Tangency	Line $y = mx+c$, Parabola $y^2=4ax$	$c = \frac{a}{m}$	Line touches parabola at exactly one real point	Keywords: 'touches parabola', 'find intercept c'
Ellipse Tangency	Line $y = mx+c$, Ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$	$c^2 = a^2 m^2 + b^2$	Line intersects ellipse at coincident boundary points	Keywords: 'tangent to ellipse', 'auxiliary circle'
Hyperbola Tangency	Line $y = mx+c$, Hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$	$c^2 = a^2 m^2 - b^2$	Line touches hyperbola branch without crossing	Keywords: 'tangent to hyperbola', 'asymptotes'
General Conic Tangency	Line $lx+my+n=0$, Conic $S=0$	$\Delta = 0$ or $CC^T = 0$	Degenerate conic condition / line touches general conic	Keywords: 'general equation', 'condition for line to touch'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the value of $k$ for which the line $y = 2x + k$ touches the parabola $y^2 = 16x$.

(A) 2(B) 4(C) 8(D) 16

📚 Method A: Formal Textbook Derivation

Step 1: Identify $a = 4$ from $y^2 = 4(4)x$. Step 2: Identify slope $m = 2$ and intercept $c = k$ from $y = 2x + k$. Step 3: Use the tangency condition $c = \frac{a}{m}$. Step 4: Calculate $k = \frac{4}{2} = 2$.

⚡ Method B: 25-Second Exam Shortcut

Apply shortcut $c = \frac{a}{m}$ directly: $k = \frac{4}{2} = 2$. Mark option (A) in 10 seconds.

💡 Tip: Always compare the given parabola with $y^2 = 4ax$ to extract $a$ correctly (here $4a = 16 \implies a = 4$).

CANONICAL PROBLEM 02 Dual-Method Breakdown

If the line $x \cos \alpha + y \sin \alpha = p$ touches the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, then the relation between $a, b, p, \alpha$ is:

(A) $p^2 = a^2 \cos^2 \alpha + b^2 \sin^2 \alpha$(B) $p^2 = a^2 \sin^2 \alpha + b^2 \cos^2 \alpha$(C) $p^2 = a^2 \cos^2 \alpha - b^2 \sin^2 \alpha$(D) $p^2 = a^2 + b^2$

📚 Method A: Formal Textbook Derivation

Step 1: Convert normal form $x \cos \alpha + y \sin \alpha = p$ to slope-intercept form $y = -\cot \alpha x + \frac{p}{\sin \alpha}$. Step 2: Here $m = -\cot \alpha$ and $c = \frac{p}{\sin \alpha}$. Step 3: Substitute into ellipse tangency condition $c^2 = a^2 m^2 + b^2$. Step 4: $\frac{p^2}{\sin^2 \alpha} = a^2 \cot^2 \alpha + b^2 \implies p^2 = a^2 \cos^2 \alpha + b^2 \sin^2 \alpha$.

⚡ Method B: 25-Second Exam Shortcut

Recall the standard parametric tangent condition for ellipses: $p^2 = a^2 \cos^2 \alpha + b^2 \sin^2 \alpha$.

💡 Tip: Standard derived result frequently asked directly as a formula in UP PGT.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: The condition that the line $lx + my + n = 0$ may touch the circle $x^2 + y^2 = a^2$ is:

A $n^2 = a^2(l^2 + m^2)$

B $n^2(l^2 + m^2) = a^2$

C $l^2 + m^2 = a^2 n^2$

D $n = a(l + m)$

Derivation & Analysis:

The perpendicular distance from the center $(0,0)$ to the line $lx + my + n = 0$ must equal the radius $a$. Thus $\frac{|n|}{\sqrt{l^2 + m^2}} = a \implies n^2 = a^2(l^2 + m^2)$.

MCQ 2: If the line $y = mx + c$ touches the parabola $y^2 = 4ax$, then the point of contact is:

A $(am^2, 2am)$

B $(\frac{a}{m^2}, \frac{2a}{m})$

C $(\frac{2a}{m^2}, \frac{a}{m})$

D $(2am, am^2)$

Derivation & Analysis:

Solving $y = mx + c$ with $y^2 = 4ax$ using $c = \frac{a}{m}$ gives the point of contact as $(\frac{a}{m^2}, \frac{2a}{m})$. ribonucleic

MCQ 3: The equation of the tangent to the circle $x^2 + y^2 - 2x - 4y - 3 = 0$ at the point $(2, 3)$ is:

A $x + y - 5 = 0$

B $x - y + 1 = 0$

C $x + 2y - 8 = 0$

D $2x + 3y - 13 = 0$

Derivation & Analysis:

Using $T = 0$: $x(2) + y(3) - 1(x + 2) - 2(y + 3) - 3 = 0 \implies 2x + 3y - x - 2 - 2y - 6 - 3 = 0 \implies x - y + 1 = 0$.

MCQ 4: The condition for the line $y = mx + c$ to be a normal to the parabola $y^2 = 4ax$ is:

A $c = -am - am^3$

B $c = am + am^3$

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

D $c = -2am - am^3$

Derivation & Analysis:

The equation of normal to $y^2 = 4ax$ in slope form is $y = mx - 2am - am^3$. Comparing with $y = mx + c$ gives $c = -2am - am^3$ (Note: sometimes expressed with parameter $m'$).

MCQ 5: If the line $lx + my = 1$ touches the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, then:

A $a^2 l^2 + b^2 m^2 = 1$

B $a^2 l^2 - b^2 m^2 = 1$

C $a^2 m^2 + b^2 l^2 = 1$

D $\frac{l^2}{a^2} + \frac{m^2}{b^2} = 1$

Derivation & Analysis:

Rewriting as $y = -\frac{l}{m}x + \frac{1}{m}$, and applying $c^2 = a^2 m^2 + b^2$, we get $\frac{1}{m^2} = a^2\left(-\frac{l}{m}\right)^2 + b^2 \implies 1 = a^2 l^2 + b^2 m^2$.

MCQ 6: The number of common tangents that can be drawn to two disjoint circles is:

A 2

B 3

C 4

D 1

Derivation & Analysis:

Two disjoint circles external to each other admit 4 common tangents: 2 direct common tangents and 2 transverse common tangents.

MCQ 7: The value of $k$ for which $3x - 4y + k = 0$ is a tangent to $x^2 + y^2 - 6x + 8y - 11 = 0$ is:

A $k = 10$ or $-30$

B $k = 5$ or $-15$

C $k = 20$ or $-40$

D $k = 0$

Derivation & Analysis:

Center of circle is $(3, -4)$, radius $r = \sqrt{9 + 16 - (-11)} = \sqrt{36} = 6$. Distance from center to line equals radius: $\frac{|3(3) - 4(-4) + k|}{\sqrt{3^2 + (-4)^2}} = 6 \implies \frac{|9 + 16 + k|}{5} = 6 \implies |25 + k| = 30 \implies 25 + k = \pm 30 \implies k = 5$ or $-55$.

MCQ 8: The chord of contact of tangents drawn from $(2, 3)$ to the circle $x^2 + y^2 = 9$ is:

A $2x + 3y = 9$

B $3x + 2y = 9$

C $2x - 3y = 9$

D $3x - 2y = 9$

Derivation & Analysis:

Chord of contact formula is $T = 0$, giving $2x + 3y = 9$.

MCQ 9: The length of the tangent drawn from any point on the circle $x^2 + y^2 + gx + fy + c_1 = 0$ to the circle $x^2 + y^2 + gx + fy + c_2 = 0$ is:

A $\sqrt{c_1 - c_2}$

B $\sqrt{c_2 - c_1}$

C $c_1 + c_2$

D $\sqrt{c_1 c_2}$

Derivation & Analysis:

The length of the tangent from a point $S_1$ to another circle is given by $\sqrt{S_1}$, which simplifies to $\sqrt{c_1 - c_2}$ when concentric or coaxial.

Q10

MCQ 10: If a line is tangent to a conic, the discriminant of the quadratic equation obtained by eliminating one variable is:

A Greater than zero

B Less than zero

C Exactly equal to zero

D Cannot be determined

Derivation & Analysis:

Tangency corresponds to coincident roots of the quadratic equation, which occurs if and only if the discriminant $\Delta = 0$.

Common questions

When a line intersects a conic, eliminating one variable yields a quadratic equation in the remaining variable. Two distinct roots mean a secant (intersecting in two points), imaginary roots mean the line misses the conic, and coincident roots ($\Delta = 0$) mean the line intersects the conic at exactly one limiting point—hence, it is a tangent.

Sign errors in standard formulas (such as confusing ellipse $c^2 = a^2m^2 + b^2$ with hyperbola $c^2 = a^2m^2 - b^2$) and failing to check whether the derivative exists at the point of tangency.

Memorize standard tangency conditions (e.g., $c = a/m$ for parabolas, $c^2 = a^2m^2 + b^2$ for ellipses) and use transformation $T = 0$ rather than expanding full algebraic lines and solving simultaneous equations from scratch.

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