JEE Main 2023 (11 Apr Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: Let $H_{n} : \frac{x ^{2}}{1 + n} - \frac{y ^{2}}{3 + n} = 1, n \in N$ . Let $k$ be the smallest even value of $n$ such that the eccentricity of $H_{k}$ is a rational number. If $l$ is the length of the latus rectum of $H_{k}$ , then $21 l$ is equal to

Q: Why did we specifically test odd perfect squares for $n+1$?

Great question! We are given that $n$ is an even natural number. If $n$ is even, then $n+1$ must be odd. Furthermore, for $e^2 = \frac{2(n+2)}{n+1}$ to be a perfect square of a rational number, the denominator $n+1$ must be a perfect square (or a multiple of one that cancels out). Testing odd perfect squares is the most efficient way to navigate the search space for $n$.

Q: Is the latus rectum formula $l = \frac{2b^2}{a}$ universal for all hyperbolas?

Yes, it is! For any standard hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the latus rectum is the chord passing through the focus perpendicular to the transverse axis. The derivation consistently yields $l = \frac{2b^2}{a}$. Always trust your standard formulas, but ensure you have correctly identified $a$ and $b$ from the given equation first.

Enter Numerical Value:

Visualized Solution

Standard Form of Hyperbola $H_{n}$

Given: $\frac{x ^{2}}{1 + n} - \frac{y ^{2}}{3 + n} = 1$
Compare with standard form: $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$
We get: $a^{2} = 1 + n$
And: $b^{2} = 3 + n$

Eccentricity Formula

Eccentricity $e$ relates $a$ and $b$ .
Formula: $e^{2} = 1 + \frac{b ^{2}}{a ^{2}}$

Substituting $a^{2}$ and $b^{2}$

Substitute $a^{2} = 1 + n$ and $b^{2} = 3 + n$ :
$e^{2} = 1 + \frac{3 + n}{1 + n}$

Simplifying $e^{2}$

$e^{2} = \frac{( 1 + n ) + ( 3 + n )}{1 + n}$
$e^{2} = \frac{2 n + 4}{n + 1}$
$e^{2} = \frac{2 ( n + 2 )}{n + 1}$

Condition for Rational Eccentricity

Given: $e$ is a rational number.
Therefore, $e^{2}$ must be the square of a rational number.
We need the smallest even natural number $n$ .

Testing Values for $n$

Since $n$ is even, $n + 1$ must be an odd number.
Let's test odd perfect squares for the denominator $n + 1$ :
If $n + 1 = 9 ⟹ n = 8 ⟹ e^{2} = \frac{2 ( 10 )}{9} = \frac{20}{9}$ (Not a square)
If $n + 1 = 25 ⟹ n = 24 ⟹ e^{2} = \frac{2 ( 26 )}{25} = \frac{52}{25}$ (Not a square)

Finding the Smallest Even $k$

Test the next odd perfect square: $n + 1 = 49$
$⟹ n = 48$
$e^{2} = \frac{2 ( 48 + 2 )}{49} = \frac{100}{49} = (\frac{10}{7})^{2}$
Smallest even value $k = 48$ .

Latus Rectum of $H_{48}$

We need the length of the latus rectum $l$ for $H_{48}$ .
Formula: $l = \frac{2 b ^{2}}{a}$
For $n = 48$ : $a^{2} = 1 + 48 = 49 ⟹ a = 7$
$b^{2} = 3 + 48 = 51$

Calculating Length $l$

Substitute $a = 7$ and $b^{2} = 51$ into $l = \frac{2 b ^{2}}{a}$ :
$l = \frac{2 ( 51 )}{7}$
$l = \frac{102}{7}$

Final Value of $21 l$

The question asks for the value of $21 l$ .
$21 l = 21 \times \frac{102}{7}$
$21 l = 3 \times 102$
$21 l = 306$

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The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Solution Diagram

Analyzing the Setup

Every great journey begins with a clear map. We are given the equation of the family of hyperbolas:

H_{n} : \frac{x ^{2}}{1 + n} - \frac{y ^{2}}{3 + n} = 1

By comparing this to the standard hyperbola equation

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1

, we immediately identify our parameters:

a^{2} = 1 + n

and

b^{2} = 3 + n

. This is our bedrock; without this identification, we are lost.

The Eccentricity Challenge

Now, we face the core of the problem: the eccentricity

e

. You know the formula by heart:

e^{2} = 1 + \frac{b ^{2}}{a ^{2}}

This formula is the heartbeat of the hyperbola. It tells us how 'open' or 'stretched' the curve is. Let us substitute our parameters:

e^{2} = 1 + \frac{3 + n}{1 + n}

With a bit of algebraic finesse, we find a common denominator:

e^{2} = \frac{( 1 + n ) + ( 3 + n )}{1 + n} = \frac{2 n + 4}{n + 1} = \frac{2 ( n + 2 )}{n + 1}

This expression is the pivot point of our entire solution.

The Number Theory Twist

Here is where the JEE Advanced spirit truly shines. We are told that

e

must be a rational number, which implies that

e^{2}

must be the square of a rational number. We are also given the constraint that

n

is an even natural number.

n

is even, then

n + 1

is odd. To make

e^{2} = \frac{2 ( n + 2 )}{n + 1}

a perfect square, we need the numerator to behave. We test values for

n + 1

that are odd perfect squares:

1. If

n + 1 = 9

, then

n = 8

, and

e^{2} = \frac{2 ( 10 )}{9} = \frac{20}{9}

(not a square). 2. If

n + 1 = 25

, then

n = 24

, and

e^{2} = \frac{2 ( 26 )}{25} = \frac{52}{25}

(not a square). 3. If

n + 1 = 49

, then

n = 48

. Substituting this, we get

e^{2} = \frac{2 ( 48 + 2 )}{49} = \frac{100}{49} = (\frac{10}{7})^{2}

We have found it! The smallest even value

n

48

Final Calculation

We are almost there. The problem asks for the length of the latus rectum

l

for

H_{48}

. The formula is:

l = \frac{2 b ^{2}}{a}

For

n = 48

, we have

a^{2} = 49

, so

a = 7

. We have

b^{2} = 3 + 48 = 51

. Thus:

l = \frac{2 ( 51 )}{7} = \frac{102}{7}

The final step is to calculate

21 l

. This is where many students rush and stumble; do not be one of them.

21 l = 21 \times \frac{102}{7} = 3 \times 102 = 306

There it is. The elegance of the cancellation is the reward for your patience. The final answer is 306.

Common Traps & FAQs

Wait, why do we reject the negative root when we talk about eccentricity?

That is a fantastic observation! By definition, the eccentricity

e

of a hyperbola is the ratio of the distance from the center to the focus to the distance from the center to the vertex. Since these are distances, they are inherently positive. Thus,

e > 1

. We only consider the positive root because eccentricity is a physical property representing the 'stretch' of the conic section, not a coordinate.

Why did we specifically test odd perfect squares for

n + 1

Great question! We are given that

n

is an even natural number. If

n

is even, then

n + 1

must be odd. Furthermore, for

e^{2} = \frac{2 ( n + 2 )}{n + 1}

to be a perfect square of a rational number, the denominator

n + 1

must be a perfect square (or a multiple of one that cancels out). Testing odd perfect squares is the most efficient way to navigate the search space for

n

Is the latus rectum formula

l = \frac{2 b ^{2}}{a}

universal for all hyperbolas?

Yes, it is! For any standard hyperbola

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1

, the latus rectum is the chord passing through the focus perpendicular to the transverse axis. The derivation consistently yields

l = \frac{2 b ^{2}}{a}

. Always trust your standard formulas, but ensure you have correctly identified

a

and

b

from the given equation first.

Enter Numerical Value:

Visualized Solution

Standard Form of Hyperbola $H_{n}$

Eccentricity Formula

Substituting $a^{2}$ and $b^{2}$

Simplifying $e^{2}$

Condition for Rational Eccentricity

Testing Values for $n$

Finding the Smallest Even $k$

Latus Rectum of $H_{48}$

Calculating Length $l$

Final Value of $21 l$

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Eccentricity Challenge

The Number Theory Twist

Final Calculation

Similar Questions

Let $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, a > b$ be an ellipse, whose eccentricity is $\frac{1}{2}$ and the length of the latus rectum is $14$ . Then the square of the eccentricity of $\frac{x ^{2}}{a ^{2}} - \frac{m y ^{2}}{b ^{2}} = 1$ is :

Let $P$ be a parabola with vertex $(2, 3)$ and directrix $2 x + y = 6$ . Let an ellipse $E : \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, a > b$ of eccentricity $\frac{1}{2}$ pass through the focus of the parabola $P$ . Then the square of the length of the latus rectum of $E$ , is

For some $θ \in (0, \frac{π}{2})$ , if the eccentricity of the hyperbola, $x^{2} - y^{2} sec^{2} θ = 10$ is $5$ times the eccentricity of the ellipse, $x^{2} sec^{2} θ + y^{2} = 5$ , then the length of the latus rectum of the ellipse, is :

Let the length of the latus rectum of an ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, (a > b)$ , be 30. If its eccentricity is the maximum value of the function $f (t) = - \frac{3}{4} + 2 t - t^{2}$ , then $(a^{2} + b^{2})$ is equal to

Let the ellipse $E : \frac{x ^{2}}{144} + \frac{y ^{2}}{169} = 1$ and the hyperbola $H : \frac{x ^{2}}{16} - \frac{y ^{2}}{λ ^{2}} = - 1$ have the same foci. If $e$ and $L$ respectively denote the eccentricity and the length of the latus rectum of $H$ , then the value of $24 (e + L)$ is :

Let the latus rectum of the hyperbola $\frac{x ^{2}}{9} - \frac{y ^{2}}{b ^{2}} = 1$ subtend an angle of $\frac{π}{3}$ at the centre of the hyperbola. If $b^{2}$ is equal to $\frac{l}{m} (1 + n)$ , where $l, m$ and $n$ are co-prime numbers, then $l^{2} + m^{2} + n^{2}$ is equal to ______

Enter Numerical Value:

Visualized Solution

Standard Form of Hyperbola .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Hn​

Eccentricity Formula

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e2

Condition for Rational Eccentricity

Testing Values for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Finding the Smallest Even .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Latus Rectum of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H48​

Calculating Length .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l

Final Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 21l

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Eccentricity Challenge

The Number Theory Twist

Final Calculation

Similar Questions

Standard Form of Hyperbola $H_{n}$

Substituting $a^{2}$ and $b^{2}$

Simplifying $e^{2}$

Testing Values for $n$

Finding the Smallest Even $k$

Latus Rectum of $H_{48}$

Calculating Length $l$

Final Value of $21 l$