How to Solve: Ellipse (Eccentricity, Standard Equations, Tangent, Normal, Latus Rectum, Director Circle) – IIT JEE Guide

How to Solve: Ellipse (Eccentricity, Standard Equations, Tangent, Normal, Latus Rectum, Director Circle) – IIT JEE Guide

Introduction

Mastering ellipses unlocks 5-10 marks in IIT JEE (Main + Advanced) every year—enough to push you from a 90th to a 99th percentile rank. Whether it’s satellite orbits, whispering galleries, or tricky coordinate geometry problems, this topic is a high-yield weapon in your exam arsenal.

What You Need To Know First

1. Coordinate Geometry Basics – Distance formula, section formula, slope of a line.
2. Conic Sections Intro – Definition of a conic (locus of points with constant ratio of distances).
3. Basic Algebra – Completing the square, quadratic equations.

Key Vocabulary

Formulas To Know

1. Standard Equation of an Ellipse

MEMORISE THIS - Horizontal Major Axis: [ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \quad (a > b) ] - (h,k) = center - a = semi-major axis - b = semi-minor axis

• Vertical Major Axis: [ \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 \quad (a > b) ]

2. Eccentricity (e)

MEMORISE THIS [ e = \sqrt{1 - \frac{b^2}{a^2}} \quad \text{(for horizontal major axis)} ] - e = eccentricity (0 < e < 1) - a = semi-major axis - b = semi-minor axis

3. Foci of the Ellipse

MEMORISE THIS - Horizontal Major Axis: [ (h \pm ae, k) ] - Vertical Major Axis: [ (h, k \pm ae) ]

4. Latus Rectum Length

MEMORISE THIS [ \text{Length} = \frac{2b^2}{a} ]

5. Equation of Tangent to Ellipse

MEMORISE THIS - At point (x₁, y₁) on the ellipse: [ \frac{x x_1}{a^2} + \frac{y y_1}{b^2} = 1 ] - Slope form (m): [ y = mx \pm \sqrt{a^2 m^2 + b^2} ]

6. Equation of Normal to Ellipse

MEMORISE THIS - At point (x₁, y₁): [ \frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2 ]

7. Director Circle

MEMORISE THIS [ x^2 + y^2 = a^2 + b^2 \quad \text{(for standard ellipse centered at origin)} ]

Step-by-Step Method

Step 1: Identify the Standard Form

• Check if the equation is in the form: [ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 ]
• If a > b, major axis is horizontal.
• If b > a, major axis is vertical.

Step 2: Extract Key Parameters

• Center (h,k)
• Semi-major axis (a)
• Semi-minor axis (b)
• Eccentricity (e) = √(1 - b²/a²)

Step 3: Find Foci

• Horizontal Major Axis: (h ± ae, k)
• Vertical Major Axis: (h, k ± ae)

Step 4: Find Latus Rectum

• Length = 2b²/a

Step 5: Find Tangent/Normal (if asked)

• Tangent at (x₁,y₁): Use (\frac{x x_1}{a^2} + \frac{y y_1}{b^2} = 1)
• Normal at (x₁,y₁): Use (\frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2)

Step 6: Director Circle (if asked)

• Equation: (x^2 + y^2 = a^2 + b^2) (for standard ellipse at origin)

Worked Examples

Example 1 – Basic (Standard Equation & Eccentricity)

Problem: Find the eccentricity of the ellipse ( \frac{x^2}{25} + \frac{y^2}{9} = 1 ).

Solution: 1. Identify a and b:
- (a^2 = 25 \implies a = 5)
- (b^2 = 9 \implies b = 3) 2. Eccentricity formula:
[
e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}
]

What we did and why: - Extracted a and b from the standard equation. - Applied the eccentricity formula directly.

Example 2 – Medium (Foci & Latus Rectum)

Problem: For the ellipse ( \frac{x^2}{16} + \frac{y^2}{7} = 1 ), find: (a) Foci (b) Length of latus rectum.

Solution: 1. Identify a and b:
- (a^2 = 16 \implies a = 4)
- (b^2 = 7 \implies b = \sqrt{7}) 2. Eccentricity:
[
e = \sqrt{1 - \frac{7}{16}} = \sqrt{\frac{9}{16}} = \frac{3}{4}
] 3. Foci:
[
( \pm ae, 0 ) = ( \pm 4 \times \frac{3}{4}, 0 ) = ( \pm 3, 0 )
] 4. Latus Rectum:
[
\text{Length} = \frac{2b^2}{a} = \frac{2 \times 7}{4} = \frac{7}{2}
]

What we did and why: - Found e first to locate foci. - Used latus rectum formula with b²/a.

Example 3 – Exam-Style (Tangent & Normal)

Problem: Find the equation of the tangent and normal to the ellipse ( \frac{x^2}{9} + \frac{y^2}{4} = 1 ) at the point ( (3/\sqrt{2}, 2/\sqrt{2}) ).

Solution: 1. Verify point lies on ellipse:
[
\frac{(3/\sqrt{2})^2}{9} + \frac{(2/\sqrt{2})^2}{4} = \frac{9/2}{9} + \frac{4/2}{4} = \frac{1}{2} + \frac{1}{2} = 1 \quad \text{(Valid)}
] 2. Tangent equation:
[
\frac{x (3/\sqrt{2})}{9} + \frac{y (2/\sqrt{2})}{4} = 1 \implies \frac{x}{3\sqrt{2}} + \frac{y}{2\sqrt{2}} = 1
]
Multiply by (6\sqrt{2}):
[
2x + 3y = 6\sqrt{2}
] 3. Normal equation:
[
\frac{9x}{3/\sqrt{2}} - \frac{4y}{2/\sqrt{2}} = 9 - 4 \implies 3\sqrt{2}x - 2\sqrt{2}y = 5
]

What we did and why: - Verified the point first (critical in exams!). - Used tangent and normal formulas with given point.

Common Mistakes

Exam Traps

1-Minute Recap (Night Before Exam)

"Listen up—this is your 60-second ellipse survival guide. First, standard equation: (\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1). If (a > b), major axis is horizontal; else, vertical. Eccentricity: (e = \sqrt{1 - b²/a²})—never forget the square root! Foci: (±ae, 0) or (0, ±ae). Latus rectum: (2b²/a). Tangent at (x₁,y₁): (\frac{x x_1}{a^2} + \frac{y y_1}{b^2} = 1). Normal: (\frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a² - b²). Director circle: (x² + y² = a² + b²). If the equation isn’t standard, divide by the constant to make RHS = 1. And always verify the point before finding tangent/normal. You’ve got this—go crush those 10 marks!