Coordinate Geometry

Standard equation of ellipse

The standard equation of an ellipse is a fundamental conic section representing the locus of a point moving such that the sum of its distances from two fixed foci is constant. This topic covers its standard.

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

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Definition and statement

Standard Equation of an Ellipse (Canonical Form)

The standard equation of an ellipse centered at the origin with axes along the coordinate axes is given by $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ where $a > b > 0$ for a horizontal ellipse, with eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}}$ ($0 < e < 1$).

📐 Geometric Interpretation

An ellipse is a closed curve symmetric about both coordinate axes, possessing two foci $S(ae, 0)$ and $S'(-ae, 0)$, two directrices $x = \pm \frac{a}{e}$, and a major axis of length $2a$ along with a minor axis of length $2b$.

⚙️ Algebraic & Structural Meaning

It represents a second-degree homogeneous and non-homogeneous algebraic curve in two variables where the discriminant $B^2 - 4AC < 0$ and the coefficients of $x^2$ and $y^2$ have the same sign but unequal magnitudes.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Non-zero denominators	`$a \neq 0, b \neq 0$ (Real constants)`	Strictly Mandatory / Necessary	Division by zero leads to undefined equations or collapses the conic into a point or pair of lines.
Inequality of semi-axes magnitude ($a \neq b$)	$a > b > 0$	Strictly Mandatory for distinct eccentricity	If $a = b$, the ellipse degenerates into a circle of radius $a$ with eccentricity $e = 0$.
Real coefficients and center at origin	`Cartesian plane $\mathbb{R}^2$`	Essential for canonical form	If the center is shifted to $(h,k)$, the standard form requires transformation to $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Eccentricity Determination

$$e = \sqrt{1 - \frac{\text{minor axis}^2}{\text{major axis}^2}} = \sqrt{1 - \frac{\text{denominator of dependent variable}}{\text{denominator of independent variable}}}$$

Rule: Always identify whether $a$ is under $x^2$ or $y^2$. The denominator under the major axis is always the larger number ($a^2$).

Application: Frequently asked in UP PGT to find $e$ directly from equations like $9x^2 + 25y^2 = 225$.

Latus Rectum Length

$$\text{Latus Rectum} = \frac{2b^2}{a}$$

Rule: Twice the square of the semi-minor axis divided by the semi-major axis. Never confuse $a$ and $b$.

Distance Between Foci

$$SS' = 2ae = 2\sqrt{a^2 - b^2}$$

Rule: The distance between the two foci is twice the product of the semi-major axis and eccentricity.

Application: Used in locus and focal property questions in TGT Mathematics.

Sum of Focal Distances

$$SP + S'P = 2a$$

Rule: The sum of the focal distances of any point on the ellipse is always equal to the length of the major axis ($2a$).

Application: Standard definition-based conceptual questions in DSSSB TGT.

Equation of Directrices

$$x = \pm \frac{a}{e}$$

Rule: For a horizontal ellipse, directrices are vertical lines given by $x = \pm \frac{a}{e}$.

Application: Directly tested in UP PGT coordinate geometry sections.

Rapid Elimination via Intercepts

$$x\text{-intercepts} = \pm a, \quad y\text{-intercepts} = \pm b$$

Rule: Instantly find where the ellipse cuts the axes by setting $y=0$ and $x=0$ to eliminate incorrect options.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Confusing Major and Minor Axes Denominators

Common Pitfall: Candidates assume $a^2$ is always under $x^2$, ignoring whether $a > b$ or $b > a$.

Counterexample: $\frac{x^2}{16} + \frac{y^2}{25} = 1$

Why it fails: Here $b^2 = 16$ and $a^2 = 25$, making the ellipse vertical (major axis along the y-axis), so $e = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$.

🛡️ Exam Defense: Always locate the larger denominator; that denominator is $a^2$, and its position determines the orientation of the major axis.

⚠️ COMMON ERROR #02

Misapplying Eccentricity Formula

Common Pitfall: Using $e = \sqrt{1 + \frac{b^2}{a^2}}$ (which is for a hyperbola) instead of $e = \sqrt{1 - \frac{b^2}{a^2}}$.

Counterexample: Ellipse eccentricity calculation

Why it fails: Results in $e > 1$, which is impossible for an ellipse ($0 < e < 1$).

🛡️ Exam Defense: Remember: Ellipse has minus ($1 - \frac{b^2}{a^2}$), Hyperbola has plus ($1 + \frac{b^2}{a^2}$). Check $e < 1$.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Horizontal Ellipse	$a > b > 0$	$_\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$	Major axis along x-axis	Denominator of $x^2$ is larger
Vertical Ellipse	$b > a > 0$	$_\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$	Major axis along y-axis	Denominator of $y^2$ is larger
Eccentricity	$0 < e < 1$	$e = \sqrt{1 - \frac{\text{minor}^2}{\text{major}^2}}$	Degree of elongation	Asked in almost every exam
Latus Rectum	$a, b > 0$	$LR = \frac{2b^2}{a}$	Chord through focus perp. to major axis	Direct numerical computation

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the eccentricity of the ellipse $9x^2 + 16y^2 = 144$.

(A) $\frac{\sqrt{7}}{4}$(B) $\frac{3}{4}$(C) $\frac{4}{5}$(D) $\frac{\sqrt{5}}{4}$

📚 Method A: Formal Textbook Derivation

Step 1: Rewrite equation in standard form by dividing by 144: $\frac{x^2}{16} + \frac{y^2}{9} = 1$. Step 2: Identify $a^2 = 16$ and $b^2 = 9$. Step 3: Apply $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{16}} = \frac{\sqrt{7}}{4}$.

⚡ Method B: 25-Second Exam Shortcut

Note $a^2 = 16, b^2 = 9$. Directly compute $e = \sqrt{1 - 9/16} = \sqrt{7}/4$.

💡 Tip: Always reduce the constant term to 1 first.

CANONICAL PROBLEM 02 Dual-Method Breakdown

If the distance between the foci of an ellipse is 6 and the length of its minor axis is 8, find its eccentricity.

(A) $\frac{3}{5}$(B) $\frac{4}{5}$(C) $\frac{3}{4}$(D) $\frac{2}{3}$

📚 Method A: Formal Textbook Derivation

Step 1: Given $2ae = 6 \implies ae = 3$. Step 2: Given minor axis $2b = 8 \implies b = 4$. Step 3: Use relation $b^2 = a^2(1-e^2) \implies b^2 = a^2 - a^2e^2 \implies 16 = a^2 - 9 \implies a^2 = 25 \implies a = 5$. Step 4: Then $e = \frac{ae}{a} = \frac{3}{5}$.

⚡ Method B: 25-Second Exam Shortcut

Use right triangle relation between $a$, $b$, and $ae$: $a^2 = b^2 + (ae)^2 = 4^2 + 3^2 = 25 \implies a = 5$. Thus $e = 3/5$.

💡 Tip: Recognize the 3-4-5 pythagorean triplet instantly in conic problems.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: The eccentricity of the conic $4x^2 + 9y^2 = 36$ is:

A $\frac{\sqrt{5}}{3}$

B $\frac{2}{3}$

C $\frac{\sqrt{5}}{2}$

D $\frac{1}{3}$

Derivation & Analysis:

Divide by 36 to get $\frac{x^2}{9} + \frac{y^2}{4} = 1$. Here $a^2 = 9, b^2 = 4$. $e = \sqrt{1 - 4/9} = \frac{\sqrt{5}}{3}$.

MCQ 2: The length of the latus rectum of the ellipse $5x^2 + 9y^2 = 45$ is:

A $\frac{10}{3}$

B $\frac{5}{3}$

C $\frac{10}{\sqrt{5}}$

D $\frac{9}{5}$

Derivation & Analysis:

Equation: $\frac{x^2}{9} + \frac{y^2}{5} = 1 \implies a^2 = 9, b^2 = 5 \implies a = 3, b = \sqrt{5}$. Latus Rectum = $\frac{2b^2}{a} = \frac{2(5)}{3} = \frac{10}{3}$.

MCQ 3: The foci of the ellipse $16x^2 + 25y^2 = 400$ are:

A $(\pm 3, 0)$

B $(0, \pm 3)$

C $(\pm 4, 0)$

D $(\pm 5, 0)$

Derivation & Analysis:

Equation in standard form: $\frac{x^2}{25} + \frac{y^2}{16} = 1$. Here $a = 5, b = 4$. $e = \sqrt{1 - 16/25} = 3/5$. Foci are $(\pm ae, 0) = (\pm 5 \times 3/5, 0) = (\pm 3, 0)$.

MCQ 4: If the eccentricity of an ellipse is $\frac{1}{2}$ and the distance between its foci is 2, then the length of the major axis is:

A 2

B 4

C 6

D 8

Derivation & Analysis:

Distance between foci = $2ae = 2 \implies ae = 1$. Given $e = 1/2$, so $a(1/2) = 1 \implies a = 2$. Major axis length = $2a = 4$.

MCQ 5: The sum of the focal distances of any point on the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ is:

A 5

B 10

C 6

D 8

Derivation & Analysis:

By definition, the sum of focal distances is equal to the major axis length $2a$. Here $a^2 = 25 \implies a = 5$, so $2a = 10$.

MCQ 6: The equation of the directrices of the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is:

A $x = \pm \frac{16}{\sqrt{7}}$

B $x = \pm \frac{\sqrt{7}}{16}$

C $y = \pm \frac{16}{\sqrt{7}}$

D $x = \pm 4$

Derivation & Analysis:

$a = 4, e = \frac{\sqrt{7}}{4}$. Directrices are $x = \pm \frac{a}{e} = \pm \frac{4}{\sqrt{7}/4} = \pm \frac{16}{\sqrt{7}}$.

MCQ 7: If the latus rectum of an ellipse is half of its major axis, then its eccentricity is:

A $\frac{1}{2}$

B $\frac{1}{\sqrt{2}}$

C $\frac{\sqrt{3}}{2}$

D $\frac{1}{3}$

Derivation & Analysis:

Given $\frac{2b^2}{a} = \frac{1}{2}(2a) = a \implies 2b^2 = a^2 \implies \frac{b^2}{a^2} = \frac{1}{2}$. Then $e = \sqrt{1 - b^2/a^2} = \sqrt{1 - 1/2} = \frac{1}{\sqrt{2}}$.

MCQ 8: Which of the following points lies on the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$?

A $(3, \frac{12}{5})$

B $(4, 3)$

C $(5, 3)$

D $(0, 5)$

Derivation & Analysis:

Substitute $(3, \frac{12}{5})$: $\frac{9}{25} + \frac{144/25}{9} = \frac{9}{25} + \frac{16}{25} = \frac{25}{25} = 1$. Satisfies the equation.

MCQ 9: The product of the perpendicular distances from the foci to any tangent of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is:

A $a^2$

B $b^2$

C $ae$

D $b$

Derivation & Analysis:

It is a standard geometric property of ellipses that the product of perpendiculars from foci to any tangent is equal to $b^2$.

Q10

MCQ 10: The eccentricity of an ellipse whose minor axis is equal to the distance between the foci is:

A $\frac{1}{\sqrt{2}}$

B $\frac{1}{2}$

C $\frac{\sqrt{3}}{2}$

D $\frac{1}{3}$

Derivation & Analysis:

Given minor axis $2b = 2ae \implies b = ae$. Substitute $b^2 = a^2e^2$. Using $b^2 = a^2(1-e^2) \implies a^2e^2 = a^2 - a^2e^2 \implies 2a^2e^2 = a^2 \implies e^2 = \frac{1}{2} \implies e = \frac{1}{\sqrt{2}}$.

Common questions

Compare the denominators of $x^2$ and $y^2$. The axis corresponding to the larger denominator is the major axis. If $a^2$ is under $x^2$, the major axis is along the x-axis.

Confusing the eccentricity formula for an ellipse ($1 - b^2/a^2$) with that of a hyperbola ($1 + b^2/a^2$), or misidentifying $a$ and $b$ when the equation is given with non-standard placement.

Memorize standard parameter relations ($b^2 = a^2(1-e^2)$, $LR = 2b^2/a$, foci coordinates $(\pm ae, 0)$) and utilize symmetry and intercept checks to eliminate wrong options instantly.

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