Why is a general result for an ellipse?

The normal line is . The denominator in the distance from the origin is . Using Cauchy-Schwarz or completing the square, its minimum is always , so (for ).

JEE MainMathematics

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Conic Sections: JEE Main Mathematics Question with Solution

If the maximum distance of normal to the ellipse

\frac{x^2}{4}+\frac{y^2}{b^2}=1

b<2

, from the origin is

1

, then the eccentricity of the ellipse is :

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

\frac{1}{2}

\frac{\sqrt{3}}{4}

\frac{\sqrt{3}}{2}

\frac{1}{\sqrt{2}}

Step-by-step solutionView

Correct answer

Using the maximum distance of a normal to the ellipse from the origin, $d_{\max} = a - b = 1$ , we find $b = 1$ and thus $e = \frac{\sqrt{3}}{2}$ .

Option analysis

Why each option works or fails

A ·

\frac{1}{2}

The student found $b/a = 1/2$ and mistook this ratio directly for the eccentricity $e$ instead of calculating $\sqrt{1 - b^2/a^2}$ . Eccentricity is given by $e = \sqrt{1 - \frac{b^2}{a^2}}$ , not the ratio of the semi-minor to semi-major axis $\frac{b}{a}$ .

B ·

\frac{\sqrt{3}}{4}

The student incorrectly simplified $\sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$ by taking an extra factor of $2$ into the denominator or computed $\frac{\sqrt{a^2 - b^2}}{a^2}$ . Ensure proper simplification of $e = \frac{\sqrt{a^2 - b^2}}{a} = \frac{\sqrt{4 - 1}}{2} = \frac{\sqrt{3}}{2}$ without squaring the denominator.

C ·

\frac{\sqrt{3}}{2}

None. This is the correct option. Correctly determined that $a - b = 1 \implies b = 1$ , yielding $e = \sqrt{1 - 1/4} = \frac{\sqrt{3}}{2}$ .

D ·

\frac{1}{\sqrt{2}}

The student used an incorrect formula for maximum distance such as $d_{\max} = \sqrt{a^2 - b^2} = 1$ , leading to $b^2 = 3$ and $e = 1/2$ , or assumed $b^2 = 2$ leading to $e = 1/\sqrt{2}$ . The maximum perpendicular distance from the origin to any normal of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $|a - b|$ , not $\sqrt{a^2 - b^2}$ .

Reviewed route

Solution

StepWorking

01given

Ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a = 2$ , $b < 2$ . Maximum distance from the origin to any normal line is $d_{\max} = 1$ .

02goal

Find the value of parameter $b$ and compute the eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}}$ .

03approach

Write the general equation of the normal at eccentric angle $\theta$ : $ax\sec\theta - by\csc\theta = a^2 - b^2$ . Express perpendicular distance from the origin $(0,0)$ to this line as a function of $\theta$ , minimize the denominator using completion of squares, equate $d_{\max} = 1$ to find $b$ , and compute $e$ .

04execute

Perpendicular distance from $(0,0)$ is: $d = \frac{|a^2 - b^2|}{\sqrt{a^2\sec^2\theta + b^2\csc^2\theta}}$ Rewrite the denominator: $a^2\sec^2\theta + b^2\csc^2\theta = a^2(1 + \tan^2\theta) + b^2(1 + \cot^2\theta) = a^2 + b^2 + (a\tan\theta - b\cot\theta)^2 + 2ab = (a+b)^2 + (a\tan\theta - b\cot\theta)^2$ The minimum value of the denominator is $\sqrt{(a+b)^2} = a+b$ , which occurs when $a\tan\theta = b\cot\theta$ . Therefore, the maximum distance is: $d_{\max} = \frac{a^2 - b^2}{a + b} = a - b$

05execute

Given $d_{\max} = 1$ and $a = 2$ , we have: $a - b = 1 \implies 2 - b = 1 \implies b = 1$ Eccentricity is: $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1^2}{2^2}} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$

✓verify

Since $b = 1 < 2$ , the condition $b < 2$ is satisfied. The major axis is along the x-axis, and $e = \frac{\sqrt{3}}{2} \in (0,1)$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why is

d_{\max} = a - b

a general result for an ellipse?

The normal line is $ax/\cos\theta - by/\sin\theta = a^2 - b^2$ . The denominator in the distance from the origin is $\sqrt{a^2\sec^2\theta + b^2\csc^2\theta}$ . Using Cauchy-Schwarz or completing the square, its minimum is always $a+b$ , so $d_{\max} = (a^2-b^2)/(a+b) = a-b$ (for $a > b$ ).

Answer

Using the maximum distance of a normal to the ellipse from the origin, $d_{\max} = a - b = 1$ , we find $b = 1$ and thus $e = \frac{\sqrt{3}}{2}$ .

Why each option works or fails

A: $\frac{1}{2}$ - The student found $b/a = 1/2$ and mistook this ratio directly for the eccentricity $e$ instead of calculating $\sqrt{1 - b^2/a^2}$ . Eccentricity is given by $e = \sqrt{1 - \frac{b^2}{a^2}}$ , not the ratio of the semi-minor to semi-major axis $\frac{b}{a}$ .
B: $\frac{\sqrt{3}}{4}$ - The student incorrectly simplified $\sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$ by taking an extra factor of $2$ into the denominator or computed $\frac{\sqrt{a^2 - b^2}}{a^2}$ . Ensure proper simplification of $e = \frac{\sqrt{a^2 - b^2}}{a} = \frac{\sqrt{4 - 1}}{2} = \frac{\sqrt{3}}{2}$ without squaring the denominator.
C · correct: $\frac{\sqrt{3}}{2}$ - None. This is the correct option. Correctly determined that $a - b = 1 \implies b = 1$ , yielding $e = \sqrt{1 - 1/4} = \frac{\sqrt{3}}{2}$ .
D: $\frac{1}{\sqrt{2}}$ - The student used an incorrect formula for maximum distance such as $d_{\max} = \sqrt{a^2 - b^2} = 1$ , leading to $b^2 = 3$ and $e = 1/2$ , or assumed $b^2 = 2$ leading to $e = 1/\sqrt{2}$ . The maximum perpendicular distance from the origin to any normal of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $|a - b|$ , not $\sqrt{a^2 - b^2}$ .

Step-by-step solution

given: Ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a = 2$ , $b < 2$ . Maximum distance from the origin to any normal line is $d_{\max} = 1$ .
goal: Find the value of parameter $b$ and compute the eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}}$ .
approach: Write the general equation of the normal at eccentric angle $\theta$ : $ax\sec\theta - by\csc\theta = a^2 - b^2$ . Express perpendicular distance from the origin $(0,0)$ to this line as a function of $\theta$ , minimize the denominator using completion of squares, equate $d_{\max} = 1$ to find $b$ , and compute $e$ .
execute: Perpendicular distance from $(0,0)$ is: $d = \frac{|a^2 - b^2|}{\sqrt{a^2\sec^2\theta + b^2\csc^2\theta}}$ Rewrite the denominator: $a^2\sec^2\theta + b^2\csc^2\theta = a^2(1 + \tan^2\theta) + b^2(1 + \cot^2\theta) = a^2 + b^2 + (a\tan\theta - b\cot\theta)^2 + 2ab = (a+b)^2 + (a\tan\theta - b\cot\theta)^2$ The minimum value of the denominator is $\sqrt{(a+b)^2} = a+b$ , which occurs when $a\tan\theta = b\cot\theta$ . Therefore, the maximum distance is: $d_{\max} = \frac{a^2 - b^2}{a + b} = a - b$
execute: Given $d_{\max} = 1$ and $a = 2$ , we have: $a - b = 1 \implies 2 - b = 1 \implies b = 1$ Eccentricity is: $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1^2}{2^2}} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$
verify: Since $b = 1 < 2$ , the condition $b < 2$ is satisfied. The major axis is along the x-axis, and $e = \frac{\sqrt{3}}{2} \in (0,1)$ .

Shortcut:

When to use it: When the standard property $d_{\max} = a - b$ for the distance of the normal to an ellipse from the origin is memorized.

given: For standard ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b$ , the maximum distance of any normal from the origin is a standard result: $d_{\max} = a - b$ . Here $a = 2$ , $d_{\max} = 1$ .

goal: Quickly obtain $b$ and evaluate $e$ .

approach: Equate $a - b = 1$ to immediately yield $b = 1$ , then use the formula $e = \sqrt{1 - b^2/a^2}$ .

execute: $2 - b = 1 \implies b = 1$ $e = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$

verify: Matches option (2).

Conic Sections: JEE Main Mathematics Question with Solution

What feels right?

Using the maximum distance of a normal to the ellipse from the origin, $d_{\max} = a - b = 1$ , we find $b = 1$ and thus $e = \frac{\sqrt{3}}{2}$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

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