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Coordinate Geometry

Standard polar equation of conics

The standard polar equation of a conic section with focus at the pole and eccentricity $e$ is expressed as $\frac{l}{r} = 1 + e \cos \theta$. This topic covers this form and its transformations.

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

Standard Polar Equation of Conics Theorem
Let the focus of a conic be at the pole $O(0,0)$ and the corresponding directrix be perpendicular to the initial line at a distance $d$ from the focus. Then the polar equation of the conic is given by $$\frac{l}{r} = 1 + e \cos(\theta - \alpha)$$ where $l$ is the semi-latus rectum, $e$ is the eccentricity, and $\alpha$ is the angle made by the axis of the conic with the initial line.

📐 Geometric Interpretation

In polar coordinates $(r, \theta)$, the pole represents the focus, $r$ is the focal distance, and $\theta$ is the vectorial angle. The parameter $e$ dictates the shape: $e=0$ (circle), $01$ (hyperbola).

⚙️ Algebraic & Structural Meaning

Converts complicated quadratic Cartesian equations $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ into a single compact reciprocal radius form $\frac{l}{r} = 1 + e \cos \theta$, greatly simplifying rotation and focal property calculations.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Focus at the Pole Fixed Origin $(r=0, \theta=0)$ Strictly Mandatory Shifting the focus away from the origin adds extra phase angles or destroys the standard $1 + e \cos \theta$ structure.
Directrix Perpendicular to Initial Line (or at angle $\alpha$) $\theta \in [0, 2\pi)$ Necessary for standard orientation If the directrix is inclined arbitrarily without angle adjustment, the cosine term must be replaced by $\cos(\theta - \alpha)$.
Non-zero Semi-latus Rectum ($l > 0$) $l \in (0, \infty)$ Strictly Mandatory If $l = 0$, the conic degenerates into a point or ray at the pole.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Semi-latus Rectum from Perihelion and Aphelion

$$l = \frac{2 r_{\min} r_{\max}}{r_{\min} + r_{\max}} = \frac{2ab^2}{a^2} \text{ (for ellipse)}$$

Rule: When closest distance $r_1$ and farthest distance $r_2$ from the focus are given, the semi-latus rectum $l$ is the harmonic mean of these distances.

Application: Frequently asked in UP PGT to find equation of ellipse/hyperbola directly from apsidal distances.
02

Reciprocal Polar Sum Property

$$ \frac{1}{r_1} + \frac{1}{r_2} = \frac{2}{l} $$

Rule: For any focal chord inclined at angle $\theta$ to the axis, the sum of the reciprocals of the segments of any focal chord is constant and equals $\frac{2}{l}$.

Application: Standard recurring formula tested in DSSSB and UP TGT vector/coordinate sections.
03

Direct Identification of Eccentricity from Equation

$$ \frac{5}{r} = 2 + 3 \cos \theta \implies \frac{5/2}{r} = 1 + \frac{3}{2}\cos \theta \implies e = \frac{3}{2} $$

Rule: Always divide the constant numerator term to make the constant term on the right side equal to $1$. The coefficient of $\cos \theta$ then directly gives $e$.

04

Tangent at Pole / Asymptotes Condition

$$ 1 + e \cos \theta_0 = 0 \implies \cos \theta_0 = -\frac{1}{e} $$

Rule: Real tangents from the pole or real asymptotes exist only if $e \ge 1$. For an ellipse ($e < 1$), $\cos \theta = -1/e$ has no real solution, confirming no real tangents from the focus.

05

Polar Equation with Directrix to the Left vs Right

$$ \frac{l}{r} = 1 \pm e \cos \theta \quad \text{and} \quad \frac{l}{r} = 1 \pm e \sin \theta $$

Rule: Use $+ e \cos \theta$ when directrix is to the right ($x = d$), $- e \cos \theta$ when to the left ($x = -d$). Use $\sin \theta$ when directrix is horizontal.

Application: Crucial for identifying correct options matching coordinate shifts.
06

Rapid Elimination via Asymptote Angle

$$ \theta = \pm \arccos\left(-\frac{1}{e}\right) $$

Rule: Eliminate options for hyperbolas where asymptotic directions fail to satisfy $\sec \theta = -e$.

Application: Saves up to 45 seconds on tough hyperbola polar problems.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

The Right-Hand Side Unity Fallacy

Common Pitfall: Given $\frac{2}{r} = 3 + 4\cos\theta$, candidates incorrectly assume $e = 4$.

Counterexample: \frac{2}{3r} = 1 + \frac{4}{3}\cos\theta \implies e = \frac{4}{3} > 1

Why it fails: The standard form requires the constant term on the RHS to be strictly $1$. Failing to normalize leads to completely wrong conic classification.

🛡️ Exam Defense: Always divide through by the constant term on the RHS before reading off $e$ or $l$.
⚠️ COMMON ERROR #02

Sign Confusion for Directrix Position

Common Pitfall: Assuming $\frac{l}{r} = 1 - e\cos\theta$ represents a directrix on the positive x-axis.

Counterexample: r = \frac{l}{1 - e\cos\theta}

Why it fails: A negative sign corresponds to a directrix located to the left of the focus ($x = -d$).

🛡️ Exam Defense: Remember: Minus sign means directrix on the negative side; plus sign means positive side.

Comparison of Conics in Polar Form

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Ellipse (Polar)$0 < e < 1$, Focus at pole$\frac{l}{r} = 1 + e \cos \theta$Closed bounded curve surrounding focus$e$ coefficient is less than 1
Parabola (Polar)$e = 1$, Focus at pole$\frac{l}{r} = 1 + \cos \theta$Open curve, symmetric about initial lineCoefficient of $\cos \theta$ is exactly 1
Hyperbola (Polar)$e > 1$, Focus at pole$\frac{l}{r} = 1 + e \cos \theta$Open two-branched curve with asymptotes$e$ coefficient is greater than 1
Rectangular Hyperbola$e = \sqrt{2}$, Focus at pole$\frac{l}{r} = 1 + \sqrt{2} \cos \theta$Asymptotes intersect at right angles$e = \sqrt{2}$ appears in problem

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the eccentricity of the conic represented by the polar equation $\frac{10}{r} = 3 + 5 \cos \theta$.

(A) $\frac{3}{5}$(B) $\frac{5}{3}$(C) $\frac{4}{3}$(D) $\frac{3}{4}$
✓ Correct Answer: (B)
📚 Method A: Formal Textbook Derivation
Step 1: Rewrite the equation in standard form by dividing by 3: $\frac{10/3}{r} = 1 + \frac{5}{3} \cos \theta$. Step 2: Compare with $\frac{l}{r} = 1 + e \cos \theta$. Step 3: Identify $e = \frac{5}{3}$. Since $e > 1$, it represents a hyperbola.
⚡ Method B: 25-Second Exam Shortcut
Directly take the ratio of the coefficient of $\cos \theta$ to the constant term on the RHS: $e = \frac{5}{3}$.
💡 Tip: Always check if the RHS constant is 1 before reading off coefficients.
CANONICAL PROBLEM 02 Dual-Method Breakdown

If $SPS'$ is a focal chord of the conic $\frac{l}{r} = 1 + e \cos \theta$ such that $SP = r_1$ and $SP' = r_2$, what is the value of $\frac{1}{r_1} + \frac{1}{r_2}$?

(A) $\frac{1}{l}$(B) $\frac{2}{l}$(C) $\frac{e}{l}$(D) $\frac{2e}{l}$
✓ Correct Answer: (B)
📚 Method A: Formal Textbook Derivation
Step 1: Let the angle of $SP$ be $\theta_1 = \alpha$. Then $SP'$ has angle $\theta_2 = \alpha + \pi$. Step 2: Write $\frac{l}{r_1} = 1 + e \cos \alpha$ and $\frac{l}{r_2} = 1 + e \cos(\alpha + \pi) = 1 - e \cos \alpha$. Step 3: Add the two equations to get $\frac{l}{r_1} + \frac{l}{r_2} = 2$, yielding $\frac{1}{r_1} + \frac{1}{r_2} = \frac{2}{l}$.
⚡ Method B: 25-Second Exam Shortcut
Standard focal chord reciprocal sum property: always equals $\frac{2}{l}$ regardless of inclination angle $\alpha$.
💡 Tip: Memorize this as a direct result for quick marking in UP PGT.

Practice questions

Choose an answer to check it and see the solution:

Q1
The polar equation $\frac{4}{r} = 1 - \cos \theta$ represents:
A An ellipse
B A parabola
C A hyperbola
D A circle
Derivation & Analysis:

Here the coefficient of $\cos \theta$ is $1$, which means eccentricity $e = 1$. Hence, the conic is a parabola.

Q2
The eccentricity of the conic $\frac{6}{r} = 2 + 2 \cos \theta$ is:
A 1
B 2
C 1/2
D 3
Derivation & Analysis:

Dividing by 2 gives $\frac{3}{r} = 1 + \cos \theta$. Thus $e = 1$, representing a parabola.

Q3
If $r = \frac{p}{1 - \cos \theta}$ is a polar equation, its Cartesian equivalent is:
A $y^2 = 2px + p^2$
B $y^2 = 2px - p^2$
C $x^2 = 2py$
D $y^2 = -2px$
Derivation & Analysis:

Substitute $r = \sqrt{x^2+y^2}$ and $r \cos \theta = x$ into $r(1 - \cos \theta) = p$ to get $\sqrt{x^2+y^2} - x = p \implies x^2+y^2 = (p+x)^2 = p^2 + 2px + x^2 \implies y^2 = 2px + p^2$.

Q4
The semi-latus rectum of the conic $\frac{5}{r} = 3 + 4 \cos \theta$ is:
A 5
B 5/3
C 3/5
D 4/3
Derivation & Analysis:

Standard form is $\frac{l/3}{r} = 1 + \frac{4}{3}\cos\theta$. Comparing $\frac{l}{3} = \frac{5}{3}$ gives $l = \frac{5}{3}$? Wait, standard form $\frac{l}{r} = 1 + e \cos\theta$ means from $\frac{5}{r} = 3(1 + \frac{4}{3}\cos\theta) \implies \frac{5/3}{r} = 1 + \frac{4}{3}\cos\theta$, so $l = \frac{5}{3}$. Let's verify: if numerator is $l$, dividing equation by 3 gives $\frac{5/3}{r}$, so $l = 5/3$. Wait, original equation $\frac{5}{r} = 3 + 4\cos\theta$, when $r$ is latus rectum $\theta=\pi/2 \implies r = 5/3$. Correct.

Q5
The polar equation of a conic with focus at pole and directrix $x = p$ is:
A $\frac{l}{r} = 1 + e \cos \theta$
B $\frac{l}{r} = 1 - e \cos \theta$
C $\frac{l}{r} = 1 + e \sin \theta$
D $\frac{l}{r} = 1 - e \sin \theta$
Derivation & Analysis:

When the directrix is to the right ($x = p > 0$), the standard polar equation takes the positive sign: $\frac{l}{r} = 1 + e \cos \theta$.

Q6
If the eccentricity $e > 1$, the polar equation $\frac{l}{r} = 1 + e \cos \theta$ represents:
A Ellipse
B Parabola
C Hyperbola
D Circle
Derivation & Analysis:

By definition, $e > 1$ corresponds to a hyperbola.

Q7
The sum of the reciprocals of the segments of any focal chord of a conic of semi-latus rectum $l$ is:
A $\frac{1}{l}$
B $\frac{2}{l}$
C $\frac{4}{l}$
D $\frac{l}{2}$
Derivation & Analysis:

By the standard focal chord property for conics, $\frac{1}{r_1} + \frac{1}{r_2} = \frac{2}{l}$.

Q8
What is the nature of the conic given by $\frac{3}{r} = 4 + 2 \cos \theta$?
A Ellipse with $e = 1/2$
B Hyperbola with $e = 2$
C Parabola with $e = 1$
D Ellipse with $e = 2/3$
Derivation & Analysis:

Dividing by 4 gives $\frac{3/4}{r} = 1 + \frac{1}{2} \cos \theta$. Here $e = 1/2 < 1$, which represents an ellipse.

Q9
The polar equation of the directrix corresponding to the focus at the pole for $\frac{l}{r} = 1 + e \cos \theta$ is:
A $r \cos \theta = \frac{l}{e}$
B $r \cos \theta = \frac{e}{l}$
C $r = \frac{l}{e}$
D $r \sin \theta = \frac{l}{e}$
Derivation & Analysis:

The directrix is given in polar coordinates by $r \cos \theta = \frac{l}{e}$, which corresponds to the Cartesian line $x = \frac{l}{e}$.

Q10
If $r = a(1 + \cos \theta)$ represents a parabola, its semi-latus rectum is:
A a
B 2a
C a/2
D 4a
Derivation & Analysis:

Rewriting $r = a(1 + \cos \theta)$ as $\frac{a}{r} = \frac{1}{1 + \cos \theta}$ or transforming to standard form $\frac{2a}{r} = 1 + \cos \theta$ shows that the semi-latus rectum $l = 2a$.

Common questions

Polar equations place the focus directly at the origin (pole), which elegantly simplifies equations involving orbital mechanics, focal properties, and central forces without cumbersome coordinate shifts.

Normalize the RHS constant to $1$ by division. The coefficient of the cosine (or sine) term is precisely the eccentricity $e$. If $e < 1$ it's an ellipse, if $e = 1$ it's a parabola, and if $e > 1$ it's a hyperbola.

Always utilize the invariant reciprocal sum formula $\frac{1}{r_1} + \frac{1}{r_2} = \frac{2}{l}$, which bypasses trigonometric angle expansions entirely.

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