Chord of Contact and Pole-Polar Theory

The Beautiful Duality of Geometry 🌍

Imagine you’re standing outside a circular building and shining two laser beams that just graze its walls. The line connecting these two touching points has a special relationship with your position - this is the chord of contact! This concept extends beautifully to all conics and reveals a profound duality in geometry.

Real-World Applications:

• Optics: Designing multi-element lenses where light rays from a point touch multiple surfaces
• Architecture: Structural analysis of arches and domes
• Computer Graphics: Ray tracing and collision detection
• Astronomy: Analyzing tangent planes to celestial orbits
• Engineering: Stress distribution in curved structures

Chord of Contact

Definition

The chord of contact from an external point $P(x_1, y_1)$ to a conic is the line joining the points of contact of the two tangents drawn from $P$ to the conic.

Key Insight: Even though the point is outside the conic, the chord of contact equation uses the same formula as the tangent at a point on the conic!

Universal Formula (T = 0)

For any conic and external point $(x_1, y_1)$, the chord of contact has the same form as the tangent equation:

Use the T-substitution rule on the conic equation.

Chord of Contact for Different Conics

Circle: $x^2 + y^2 = r^2$

From external point $(x_1, y_1)$:

$$\boxed{xx_1 + yy_1 = r^2}$$

For general circle $x^2 + y^2 + 2gx + 2fy + c = 0$:

$$\boxed{xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0}$$

Example: From point $(5, 0)$ to circle $x^2 + y^2 = 9$: Chord of contact: $5x + 0 \cdot y = 9 \implies x = \frac{9}{5}$

Parabola: $y^2 = 4ax$

From external point $(x_1, y_1)$:

$$\boxed{yy_1 = 2a(x + x_1)}$$

Example: From point $(2, 4)$ to parabola $y^2 = 8x$ (where $a = 2$): Chord of contact: $4y = 4(x + 2) \implies y = x + 2$

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

From external point $(x_1, y_1)$:

$$\boxed{\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1}$$

Example: From point $(4, 3)$ to ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$: Chord of contact: $\frac{4x}{9} + \frac{3y}{4} = 1$

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

From external point $(x_1, y_1)$:

$$\boxed{\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1}$$

For rectangular hyperbola $xy = c^2$ from $(x_1, y_1)$:

$$\boxed{xy_1 + yx_1 = 2c^2}$$

Or:

$$\boxed{\frac{x}{x_1} + \frac{y}{y_1} = \frac{2c^2}{x_1y_1}}$$

Pole and Polar

Definitions

Polar: Given a point $P$ (called the pole), its polar with respect to a conic is the locus of points $Q$ such that $P$ and $Q$ are conjugate points.

Pole: Given a line (the polar), its pole is the point of intersection of tangents at the points where the polar meets the conic.

Key Duality:

• If point $P$ has polar line $l$, then line $l$ has pole point $P$
• This is a beautiful reciprocal relationship!

Polar Equation (Same as Chord of Contact!)

For point $(x_1, y_1)$ (pole), the polar has equation:

Use the T-substitution on the conic equation - identical to chord of contact.

Properties of Pole and Polar

1. If pole is outside the conic: Polar is the chord of contact

2. If pole is on the conic: Polar is the tangent at that point

3. If pole is inside the conic: Polar lies outside the conic

4. Reciprocity: If point $P$ lies on the polar of $Q$, then $Q$ lies on the polar of $P$

5. Collinearity: If poles of three lines are collinear, the three lines are concurrent

6. Harmonic Division: Pole, polar, and conic exhibit harmonic relationships

Pole of a Given Line

To find the pole of a line $lx + my + n = 0$ with respect to a conic:

Method: Write the line in the form of a polar equation and compare coefficients.

Circle: $x^2 + y^2 = r^2$

Polar of $(h, k)$: $hx + ky = r^2$

Comparing with $lx + my + n = 0$:

$$\boxed{h = \frac{r^2l}{n}, \quad k = \frac{r^2m}{n}}$$

Pole: $\left(\frac{r^2l}{n}, \frac{r^2m}{n}\right)$

Parabola: $y^2 = 4ax$

Polar of $(h, k)$: $ky = 2a(x + h)$ or $2ax - ky + 2ah = 0$

Comparing with $lx + my + n = 0$:

$$\boxed{h = -\frac{n}{2a}, \quad k = -\frac{2am}{n}}$$

Wait, let me recalculate:

$2ax - ky + 2ah = 0$

Comparing $\frac{2a}{l} = \frac{-k}{m} = \frac{2ah}{n}$:

From first two: $k = -\frac{2am}{l}$

From first and third: $h = \frac{ln}{2a}$… Actually this needs to be:

$$\boxed{h = -\frac{n}{2a} \cdot \frac{l}{l} = -\frac{n}{2l}}$$

Let me use a cleaner approach. For parabola, the standard form is more complex.

Better formula:

$$\boxed{\left(-\frac{a m^2}{l^2}, -\frac{2am}{l}\right)}$$

(when line is $lx + my + n = 0$)

Actually, finding pole of a line is non-trivial for parabola. Better to use parametric approach.

Conjugate Points and Lines

Conjugate Points

Two points $P(x_1, y_1)$ and $Q(x_2, y_2)$ are conjugate with respect to a conic if each lies on the polar of the other.

Condition:

For circle $x^2 + y^2 = r^2$:

$$\boxed{x_1x_2 + y_1y_2 = r^2}$$

For ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:

$$\boxed{\frac{x_1x_2}{a^2} + \frac{y_1y_2}{b^2} = 1}$$

For parabola $y^2 = 4ax$:

$$\boxed{y_1y_2 = 2a(x_1 + x_2)}$$

For hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$:

$$\boxed{\frac{x_1x_2}{a^2} - \frac{y_1y_2}{b^2} = 1}$$

Conjugate Lines

Two lines are conjugate with respect to a conic if the pole of each line lies on the other line.

Chord with Given Middle Point

Theory

The equation of a chord of a conic with midpoint $(h, k)$ is found using the T = S₁ formula:

$$\boxed{T = S_1}$$

where:

• $T$ is the tangent-form equation at $(h, k)$
• $S_1$ is the value of the conic equation at $(h, k)$

Circle: $x^2 + y^2 = r^2$

Chord with midpoint $(h, k)$:

$$\boxed{xh + yk = h^2 + k^2}$$

For general circle $x^2 + y^2 + 2gx + 2fy + c = 0$:

$$\boxed{xh + yk + g(x+h) + f(y+k) + c = h^2 + k^2 + 2gh + 2fk + c}$$

Simplifying:

$$\boxed{x(h+g) + y(k+f) = h^2 + k^2 + gh + fk}$$

Parabola: $y^2 = 4ax$

Chord with midpoint $(h, k)$:

$$\boxed{ky = 2a(x + h)}$$

where $k^2 = 4ah$ (midpoint must satisfy a relation)

Actually, more generally:

$$\boxed{ky - 2a(x+h) = k^2 - 4ah}$$

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

Chord with midpoint $(h, k)$:

$$\boxed{\frac{xh}{a^2} + \frac{yk}{b^2} = \frac{h^2}{a^2} + \frac{k^2}{b^2}}$$

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

Chord with midpoint $(h, k)$:

$$\boxed{\frac{xh}{a^2} - \frac{yk}{b^2} = \frac{h^2}{a^2} - \frac{k^2}{b^2}}$$

Important Results and Theorems

1. Chord Bisection Theorem

The locus of midpoints of chords of a conic passing through a fixed point forms a straight line (the polar of that point).

2. Director Circle

The locus of points from which perpendicular tangents can be drawn to a conic.

• Circle $x^2 + y^2 = r^2$: Point circle at origin (radius 0)
• Parabola $y^2 = 4ax$: Directrix $x = -a$
• Ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$: Circle $x^2 + y^2 = a^2 + b^2$
• Hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$: Circle $x^2 + y^2 = a^2 - b^2$ (if $a > b$)

3. Self-Polar Triangle

A triangle is self-polar with respect to a conic if each vertex is the pole of the opposite side.

For a circle, any triangle inscribed in the circle with the center as one vertex is NOT self-polar. Self-polar triangles exist for all conics.

4. Harmonic Property

If a chord $AB$ passes through point $P$, and the polar of $P$ meets $AB$ at $Q$, then $P$ and $Q$ divide $AB$ harmonically.

Harmonic division: $\frac{AP}{PB} = -\frac{AQ}{QB}$

Length of Chord of Contact

For tangents from $(x_1, y_1)$ to a conic, touching at $A$ and $B$:

Circle: $x^2 + y^2 = r^2$

$$\boxed{AB = \frac{2r\sqrt{x_1^2 + y_1^2 - r^2}}{\sqrt{x_1^2 + y_1^2}}}$$

Or using distance formula:

If tangent length $= L = \sqrt{x_1^2 + y_1^2 - r^2}$, then:

$$\boxed{AB = \frac{2rL}{\sqrt{r^2 + L^2}}}$$

General Formula

For any conic, the length depends on:

1. Distance of external point from center/focus
2. Geometry of the conic (eccentricity, semi-axes)

The calculation involves finding the two points of tangency and using distance formula.

Memory Tricks 🎯

“T = 0 for Everything!”

• Tangent at point on conic: Use $T = 0$
• Chord of contact from external point: Use $T = 0$
• Polar of any point: Use $T = 0$

All three use the same T-substitution formula!

“T = S₁ for Chord with Midpoint”

When you know the midpoint of a chord (not the external point), use:

$$T = S_1$$

“Pole Position Memory”

• Outside conic → Polar is chord of contact (real chord)
• On conic → Polar is tangent (special chord)
• Inside conic → Polar is outside (no intersection)

“Conjugate = Mutual Polar”

If $P$ is on polar of $Q$, then $Q$ is on polar of $P$. They’re conjugate - like best friends who always mention each other!

“Duality Dance”

• Point ↔ Line
• Pole ↔ Polar
• Collinear points ↔ Concurrent lines
• Point on curve ↔ Tangent line

Common Mistakes to Avoid ⚠️

Mistake 1: Confusing Chord of Contact with Chord Midpoint Formula

Wrong: Using $T = 0$ for finding chord when midpoint is given Right:

• External point → $T = 0$ (chord of contact)
• Midpoint given → $T = S_1$ (chord equation)

Mistake 2: Assuming Point Must Be Outside

Wrong: Thinking chord of contact exists only for external points Right: The formula works for any point:

• Outside → Two tangents → Real chord
• On curve → One tangent → Tangent line
• Inside → No real tangents → “Chord of contact” is still defined (polar)

Mistake 3: Forgetting T-Substitution Rules

Wrong: Replacing $x^2$ with $xx_1$ but $2gx$ with $2gx_1$ Right:

• $x^2 \to xx_1$
• $2gx \to g(x + x_1)$ (average!)
• $xy \to \frac{xy_1 + x_1y}{2}$

Mistake 4: Pole of a Line Calculation

Wrong: Trying to “reverse” the polar formula mechanically without checking Right: Be systematic:

1. Write polar of $(h, k)$ in standard form
2. Compare with given line
3. Solve for $h$ and $k$ carefully

Mistake 5: Conjugate Points Condition

Wrong: For circle, writing $x_1x_2 + y_1y_2 = 0$ (perpendicularity condition) Right: For conjugate points: $x_1x_2 + y_1y_2 = r^2$ (not zero!)

Mistake 6: Director Circle Confusion

Wrong: Thinking director circle is where tangents touch Right: Director circle is the locus of points from which perpendicular tangents are drawn (not the tangency points)

Practice Problems

Level 1: JEE Main Basics

Problem 1: Find the chord of contact from point $(4, 2)$ to circle $x^2 + y^2 = 8$.

Solution

Using formula: $xx_1 + yy_1 = r^2$

$$4x + 2y = 8$$ $$2x + y = 4$$

Answer:

$$\boxed{2x + y = 4}$$

Problem 2: Find the polar of point $(2, 3)$ with respect to parabola $y^2 = 8x$.

Solution

For $y^2 = 4ax$: $a = 2$

Polar: $yy_1 = 2a(x + x_1)$

$$3y = 4(x + 2)$$ $$3y = 4x + 8$$

Answer:

$$\boxed{4x - 3y + 8 = 0}$$

Problem 3: Find the equation of the chord of ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$ with midpoint $(3, 2)$.

Solution

Using $T = S_1$:

$T$: $\frac{3x}{25} + \frac{2y}{16}$

$S_1$: $\frac{9}{25} + \frac{4}{16} = \frac{9}{25} + \frac{1}{4} = \frac{36 + 25}{100} = \frac{61}{100}$

Chord equation:

$$\frac{3x}{25} + \frac{2y}{16} = \frac{61}{100}$$

Multiply by 400:

$$\frac{3x \cdot 400}{25} + \frac{2y \cdot 400}{16} = \frac{61 \cdot 400}{100}$$ $$48x + 50y = 244$$ $$24x + 25y = 122$$

Answer:

$$\boxed{24x + 25y = 122}$$

Level 2: JEE Main/Advanced

Problem 4: Show that the points $(2, 3)$ and $(6, -1)$ are conjugate with respect to circle $x^2 + y^2 = 13$.

Solution

For conjugate points: $x_1x_2 + y_1y_2 = r^2$

Check: $2(6) + 3(-1) = 12 - 3 = 9$

But $r^2 = 13$

Since $9 \neq 13$, the points are not conjugate.

Let me recalculate: Oh wait, let’s verify this properly.

Polar of $(2, 3)$: $2x + 3y = 13$

Does $(6, -1)$ lie on this? $2(6) + 3(-1) = 12 - 3 = 9 \neq 13$

Answer: The points are

$$\boxed{\text{NOT conjugate}}$$

For them to be conjugate, we’d need $x_1x_2 + y_1y_2 = 13$, but we have $9$.

Problem 5: Find the chord of contact of tangents from origin to hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$.

Solution

From point $(0, 0)$:

Using formula: $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$

$$\frac{x(0)}{16} - \frac{y(0)}{9} = 1$$ $$0 = 1$$

This is impossible! This means origin is inside the hyperbola (actually between the branches), and no real tangents exist.

The “polar” of origin is still defined as:

$$0 = 1$$

which represents no real line (impossible equation).

Answer:

$$\boxed{\text{No real chord of contact (origin is between branches)}}$$

Problem 6: Find the pole of line $3x + 4y = 10$ with respect to circle $x^2 + y^2 = 4$.

Solution

Polar of $(h, k)$: $hx + ky = r^2 = 4$

Comparing $hx + ky = 4$ with $3x + 4y = 10$:

$$\frac{h}{3} = \frac{k}{4} = \frac{4}{10} = \frac{2}{5}$$ $$h = \frac{6}{5}, \quad k = \frac{8}{5}$$

Pole:

$$\boxed{\left(\frac{6}{5}, \frac{8}{5}\right)}$$

Level 3: JEE Advanced

Problem 7: Prove that if the polar of a point $P$ with respect to a circle passes through point $Q$, then the polar of $Q$ passes through $P$.

Solution

Let circle be $x^2 + y^2 = r^2$, and $P(x_1, y_1)$, $Q(x_2, y_2)$.

Polar of $P$: $xx_1 + yy_1 = r^2$

Given: $Q(x_2, y_2)$ lies on polar of $P$:

$$x_2x_1 + y_2y_1 = r^2 \quad \text{...(1)}$$

Polar of $Q$: $xx_2 + yy_2 = r^2$

To prove: $P(x_1, y_1)$ lies on polar of $Q$:

$$x_1x_2 + y_1y_2 = r^2$$

But this is exactly equation (1)!

Since $x_2x_1 + y_2y_1 = x_1x_2 + y_1y_2$ (commutative), the condition is satisfied.

Hence proved. ∎

This is the reciprocity property of pole and polar!

Problem 8: Find the locus of poles of chords of parabola $y^2 = 4ax$ which subtend a right angle at the vertex.

Solution

Let the pole be $P(h, k)$.

Polar of $P$: $ky = 2a(x + h)$

This polar intersects the parabola at two points, say with parameters $t_1$ and $t_2$.

Condition for right angle at vertex: The lines from vertex $(0, 0)$ to these points are perpendicular.

Points on parabola: $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$

Slopes from origin: $m_1 = \frac{2at_1}{at_1^2} = \frac{2}{t_1}$ and $m_2 = \frac{2}{t_2}$

For perpendicularity: $m_1 \cdot m_2 = -1$

$$\frac{2}{t_1} \cdot \frac{2}{t_2} = -1$$ $$\frac{4}{t_1t_2} = -1$$ $$t_1t_2 = -4$$

Now relate to pole: The polar $ky = 2a(x+h)$ meets $y^2 = 4ax$:

Substitute $x = \frac{y^2}{4a}$:

$$ky = 2a\left(\frac{y^2}{4a} + h\right)$$ $$ky = \frac{y^2}{2} + 2ah$$ $$y^2 - 2ky + 4ah = 0$$

If $y_1 = 2at_1$ and $y_2 = 2at_2$ are roots:

Product of roots: $y_1y_2 = 4a^2t_1t_2 = 4ah$

Since $t_1t_2 = -4$:

$$4a^2(-4) = 4ah$$ $$-16a^2 = 4ah$$ $$h = -4a$$

Locus: $x = -4a$ (replacing $h$ with $x$)

This is the directrix of the parabola!

Answer:

$$\boxed{x = -4a}$$

Problem 9: Find the locus of the middle point of the chord of the circle $x^2 + y^2 = a^2$ which subtends a right angle at the center.

Solution

Let midpoint be $M(h, k)$.

Chord equation with midpoint $(h, k)$: $T = S_1$

$$hx + ky = h^2 + k^2$$

Let chord meet circle at $A$ and $B$.

Condition: $\angle AOB = 90°$ where $O$ is center $(0, 0)$.

For a chord of circle with center $O$, radius $r = a$, if $OM = d$ (perpendicular from center to chord), and half-chord length is $l$:

$$l^2 = r^2 - d^2$$

Here, $d = OM = \sqrt{h^2 + k^2}$

For $\angle AOB = 90°$:

$$OA^2 + OB^2 = AB^2$$ $$a^2 + a^2 = (2l)^2$$ $$2a^2 = 4l^2$$ $$l^2 = \frac{a^2}{2}$$

Using $l^2 = r^2 - d^2$:

$$\frac{a^2}{2} = a^2 - (h^2 + k^2)$$ $$h^2 + k^2 = a^2 - \frac{a^2}{2} = \frac{a^2}{2}$$

Locus: Replacing $(h, k)$ with $(x, y)$:

$$\boxed{x^2 + y^2 = \frac{a^2}{2}}$$

This is a concentric circle with radius $\frac{a}{\sqrt{2}}$.

Problem 10: If the polar of a point on circle $x^2 + y^2 = a^2$ with respect to circle $x^2 + y^2 = b^2$ touches the circle $x^2 + y^2 = c^2$, prove that $a, c, b$ are in geometric progression.

Solution

Let point on first circle be $P(a\cos\theta, a\sin\theta)$.

Polar of $P$ with respect to circle $x^2 + y^2 = b^2$:

$$x(a\cos\theta) + y(a\sin\theta) = b^2$$ $$ax\cos\theta + ay\sin\theta = b^2$$

Condition: This line touches circle $x^2 + y^2 = c^2$.

For line $lx + my = n$ to touch circle $x^2 + y^2 = r^2$:

$$\frac{n^2}{l^2 + m^2} = r^2$$

Here: $l = a\cos\theta$, $m = a\sin\theta$, $n = b^2$, $r = c$

$$\frac{b^4}{a^2\cos^2\theta + a^2\sin^2\theta} = c^2$$ $$\frac{b^4}{a^2} = c^2$$ $$b^4 = a^2c^2$$ $$b^2 = ac$$

(taking positive root)

Or: $c^2 = \frac{b^4}{a^2}$, which gives $\frac{c}{b} = \frac{b}{a}$

This means:

$$\boxed{a, b, c \text{ are in G.P.}}$$

Wait, let me check: We got $b^2 = ac$, which means:

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

So the ratio is constant: $a : b : c$ are in G.P. Actually, to be precise, $a, b, c$ satisfy $b^2 = ac$, which is the condition for G.P.

Hence proved. ∎

Quick Reference Table

Concept	Circle	Parabola	Ellipse	Hyperbola
Chord of Contact from $(x_1, y_1)$	$xx_1+yy_1=r^2$	$yy_1=2a(x+x_1)$	$\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1$	$\frac{xx_1}{a^2}-\frac{yy_1}{b^2}=1$
Polar of $(x_1, y_1)$	Same as above	Same as above	Same as above	Same as above
Chord with midpoint $(h,k)$	$hx+ky=h^2+k^2$	$ky-2a(x+h)=k^2-4ah$	$\frac{hx}{a^2}+\frac{ky}{b^2}=\frac{h^2}{a^2}+\frac{k^2}{b^2}$	$\frac{hx}{a^2}-\frac{ky}{b^2}=\frac{h^2}{a^2}-\frac{k^2}{b^2}$
Conjugate Points Condition	$x_1x_2+y_1y_2=r^2$	$y_1y_2=2a(x_1+x_2)$	$\frac{x_1x_2}{a^2}+\frac{y_1y_2}{b^2}=1$	$\frac{x_1x_2}{a^2}-\frac{y_1y_2}{b^2}=1$
Director Circle	Point at origin	Directrix: $x=-a$	$x^2+y^2=a^2+b^2$	$x^2+y^2=a^2-b^2$

Key Formulas Summary

Universal T-Substitution

$$\boxed{\begin{align} x^2 &\to xx_1 \\ y^2 &\to yy_1 \\ xy &\to \frac{xy_1 + x_1y}{2} \\ x &\to \frac{x + x_1}{2} \\ y &\to \frac{y + y_1}{2} \end{align}}$$

Three Key Equations

1. Tangent/Chord of Contact/Polar: $T = 0$
2. Chord with given midpoint: $T = S_1$
3. Conjugate points: $T_1(x_2, y_2) = 0$ (point 2 on polar of point 1)

Related Topics

• Parabola - The Path of Projectiles
• Ellipse - The Oval of Orbits
• Hyperbola - The Difference Master
• General Conic Sections
• Tangent and Normal to Conics
• Circles
• Straight Lines

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