---
title: "Condition for lx+my+n=0 to be normal to ellipse"
subject: "Mathematics I"
chapter: 3
key: "semester_1/mathematics_i:ellipse-normal-condition-lx-my-n"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/ellipse-normal-condition-lx-my-n
site: BCT Engineering
---

# Condition for lx+my+n=0 to be normal to ellipse

Mathematics I · Chapter 3: Plane Analytic Geometry

- Asked in: 2072 Kartik; 2071 Chaitra (asked 2 times)
- Video: [The line lx+my+n=0 is a normal to the ellipse x^2/a^2+y^2/b^2=1 . then](https://www.youtube.com/watch?v=BnLxtowjoh4) — JEE MATHS With LOV SIR 
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/ellipse-normal-condition-lx-my-n
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-3

## Question

Find the condition that the line lx + my + n = 0 may be a normal to the ellipse x²/a² + y²/b² = 1. (Show that a²/l² + b²/m² = (a²−b²)²/n².)

## Short answer

The line lx+my+n=0 is normal to x²/a²+y²/b²=1 if a²/l² + b²/m² = (a²−b²)²/n². This follows by equating coefficients with the normal at (a cosφ, b sinφ) and using cos²φ+sin²φ=1.

## Answer

Use the parametric point P(a cosφ, b sinφ) on the ellipse x²/a² + y²/b² = 1.

#### Normal at P
The tangent at P is x cosφ/a + y sinφ/b = 1, with slope −(b cosφ)/(a sinφ). The normal is perpendicular to it and passes through P:

**ax/cosφ − by/sinφ = a² − b²**

#### Comparing with lx + my + n = 0
Write the given line as lx + my = −n. Two equations represent the same line if their coefficients are proportional:

(a/cosφ)/l = (−b/sinφ)/m = (a² − b²)/(−n)

From this:

cosφ = −an / [l(a² − b²)]
sinφ = bn / [m(a² − b²)]

#### Eliminate φ
Using cos²φ + sin²φ = 1:

a²n²/[l²(a² − b²)²] + b²n²/[m²(a² − b²)²] = 1

Multiplying by (a² − b²)²/n²:

**a²/l² + b²/m² = (a² − b²)²/n²**

This is the required condition for lx + my + n = 0 to be a normal to the ellipse.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the normal at (a cosφ, b sinφ) of x^2/a^2 + y^2/b^2 = 1?
   A. ax/cosφ + by/sinφ = a^2 + b^2
   B. x cosφ/a + y sinφ/b = 1
   C. ax/cosφ − by/sinφ = a^2 − b^2
   D. bx/cosφ − ay/sinφ = a^2 − b^2
   Answer: C. It is the line through P perpendicular to the tangent. (Key point: Normal at P)

2. How is the condition obtained?
   A. Equate the slope of the line to zero
   B. Compare the normal with lx + my + n = 0 (proportional coefficients) to find cosφ and sinφ
   C. Substitute the line into the ellipse and set the discriminant zero
   D. Put φ = 0
   Answer: B. Matching coefficients gives cosφ = −an/[l(a^2−b^2)] and sinφ = bn/[m(a^2−b^2)]. (Key point: Comparing coefficients)

3. Which identity eliminates φ?
   A. cos^2φ + sin^2φ = 1
   B. tanφ = sinφ/cosφ
   C. sec^2φ − tan^2φ = 1
   D. sin2φ = 2 sinφ cosφ
   Answer: A. Squaring and adding the expressions gives the condition. (Key point: Eliminating φ)

4. What is the final condition?
   A. a^2l^2 + b^2m^2 = n^2
   B. a^2/l^2 − b^2/m^2 = (a^2+b^2)^2/n^2
   C. l^2/a^2 + m^2/b^2 = n^2
   D. a^2/l^2 + b^2/m^2 = (a^2 − b^2)^2/n^2
   Answer: D. Multiplying by (a^2−b^2)^2/n^2 gives the required normal condition; note this differs from the tangent condition. (Key point: Result)

## Related in Mathematics I

- [Condition for lx+my+n=0 to be tangent to ellipse](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aellipse-tangent-condition-lx-my-n) — Mathematics I, chapter 3
- [Condition for x cosα + y sinα = p to touch ellipse](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aellipse-tangent-condition-xcos-ysin) — Mathematics I, chapter 3