JEE Main 2022 (28 June Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: Let $a > 0, b > 0$ . Let $e$ and $ℓ$ respectively be the eccentricity and length of the latus rectum of the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ . Let $e^{'}$ and $ℓ^{'}$ respectively be the eccentricity and length of the latus rectum of its conjugate hyperbola. If $e^{2} = \frac{11}{14} ℓ$ and $(e^{'})^{2} = \frac{11}{8} ℓ^{'}$ , then the value of $77 a + 44 b$ is equal to:

Q: Wait, why do we reject the negative root or why don't we solve for $e$ directly?

Great question! In JEE Advanced problems, solving for $e$ directly often introduces unnecessary square roots and complicates the algebra. By keeping the equations in terms of $e^2$ and $(e')^2$, we work with rational expressions that are much easier to manipulate. We don't reject roots here because $a$ and $b$ are defined as positive lengths, so we are strictly in the first quadrant of the parameter space.

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the symmetry. The hyperbola

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

and its conjugate

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

are intimately linked. Notice that the term

a^{2} + b^{2}

appears in the eccentricity formula for both curves. This is the hidden geometric reality you must exploit.

Hint 2

Hint 2: Recall your toolkit. The eccentricity squared is

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

and the latus rectum is

ℓ = \frac{2 b ^{2}}{a}

. For the conjugate, simply swap

a

and

b

. Do not try to solve for

e

e^{'}

individually; keep them as squares to simplify the algebra.

Hint 3

Hint 3: The Execution. Set up the two equations:

\frac{a ^{2} + b ^{2}}{a} = \frac{11 b ^{2}}{7}

and

\frac{a ^{2} + b ^{2}}{b} = \frac{11 a ^{2}}{4}

. The most efficient path forward is to divide the first equation by the second. This will cause the complex

a^{2} + b^{2}

term to vanish, leaving you with a simple ratio between

a

and

b

Visualized Solution

Introduction to Hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$

Given hyperbola: $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$
Conjugate hyperbola: $\frac{y ^{2}}{b ^{2}} - \frac{x ^{2}}{a ^{2}} = 1$
Objective: Find $77 a + 44 b$ using given conditions.

Properties of Hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$

For hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ :
Eccentricity squared: $e^{2} = 1 + \frac{b ^{2}}{a ^{2}} = \frac{a ^{2} + b ^{2}}{a ^{2}}$
Length of latus rectum: $ℓ = \frac{2 b ^{2}}{a}$

Applying the First Condition $e^{2} = \frac{11}{14} ℓ$

Given condition: $e^{2} = \frac{11}{14} ℓ$
Substitute the formulas: $\frac{a ^{2} + b ^{2}}{a ^{2}} = \frac{11}{14} (\frac{2 b ^{2}}{a})$

Simplifying the First Equation

Cancel $a$ from denominators (since $a > 0$ )
Cancel $2$ with $14$ to get $7$
Simplified equation: $\frac{a ^{2} + b ^{2}}{a} = \frac{11 b ^{2}}{7}$ ... (1)

Properties of Conjugate Hyperbola $\frac{y ^{2}}{b ^{2}} - \frac{x ^{2}}{a ^{2}} = 1$

For conjugate hyperbola $\frac{y ^{2}}{b ^{2}} - \frac{x ^{2}}{a ^{2}} = 1$ :
Eccentricity squared: $(e^{'})^{2} = 1 + \frac{a ^{2}}{b ^{2}} = \frac{a ^{2} + b ^{2}}{b ^{2}}$
Length of latus rectum: $ℓ^{'} = \frac{2 a ^{2}}{b}$

Applying the Second Condition $(e^{'})^{2} = \frac{11}{8} ℓ^{'}$

Given condition: $(e^{'})^{2} = \frac{11}{8} ℓ^{'}$
Substitute the formulas: $\frac{a ^{2} + b ^{2}}{b ^{2}} = \frac{11}{8} (\frac{2 a ^{2}}{b})$

Simplifying the Second Equation

Cancel $b$ from denominators (since $b > 0$ )
Cancel $2$ with $8$ to get $4$
Simplified equation: $\frac{a ^{2} + b ^{2}}{b} = \frac{11 a ^{2}}{4}$ ... (2)

Dividing Equation (1) by (2)

Divide (1) by (2): $\frac{\frac{a ^{2} + b ^{2}}{a}}{\frac{a ^{2} + b ^{2}}{b}} = \frac{\frac{11 b ^{2}}{7}}{\frac{11 a ^{2}}{4}}$
The term $(a^{2} + b^{2})$ cancels out.
Left side becomes $\frac{b}{a}$
Right side becomes $\frac{4 b ^{2}}{7 a ^{2}}$

Finding the Ratio of $a$ and $b$

We have: $\frac{b}{a} = \frac{4 b ^{2}}{7 a ^{2}}$
Since $a, b > 0$ , divide both sides by $\frac{b}{a}$
$1 = \frac{4 b}{7 a}$
Rearranging gives: $7 a = 4 b ⟹ b = \frac{7 a}{4}$

Substituting $b = \frac{7 a}{4}$ into Equation (2)

Recall Equation (2): $\frac{a ^{2} + b ^{2}}{b} = \frac{11 a ^{2}}{4}$
Substitute $b = \frac{7 a}{4}$ :
$\frac{a ^{2} + ( \frac{7 a}{4} ) ^{2}}{\frac{7 a}{4}} = \frac{11 a ^{2}}{4}$

Solving for $a$

Numerator: $a^{2} + \frac{49 a ^{2}}{16} = \frac{65 a ^{2}}{16}$
Equation becomes: $\frac{\frac{65 a ^{2}}{16}}{\frac{7 a}{4}} = \frac{11 a ^{2}}{4}$
Simplify left side: $\frac{65 a}{28} = \frac{11 a ^{2}}{4}$
Cancel $a$ and solve: $a = \frac{65}{77}$

Calculating the Value of $b$

We know $b = \frac{7 a}{4}$
Substitute $a = \frac{65}{77}$ :
$b = \frac{7}{4} \times \frac{65}{77}$
Cancel $7$ with $77$ to get $11$ in denominator.
$b = \frac{65}{44}$

Final Calculation: $77 a + 44 b$

We need to find: $77 a + 44 b$
Substitute $a = \frac{65}{77}$ and $b = \frac{65}{44}$ :
$77 (\frac{65}{77}) + 44 (\frac{65}{44})$
$= 65 + 65 = 130$
Final Answer: $130$

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

Solution Diagram

The Anatomy of the Hyperbola

Let us first dissect the blue hyperbola defined by the equation

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

. Its eccentricity squared is given by:

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

The length of its latus rectum, the chord passing through the focus, is defined as:

ℓ = \frac{2 b ^{2}}{a}

The problem provides a bridge between these two properties:

e^{2} = \frac{11}{14} ℓ

. By substituting our expressions, we obtain:

\frac{a ^{2} + b ^{2}}{a ^{2}} = \frac{11}{14} (\frac{2 b ^{2}}{a})

With a steady hand, we simplify this expression. Canceling an

a

from the denominators and reducing the fraction, we arrive at our first pillar:

\frac{a ^{2} + b ^{2}}{a} = \frac{11 b ^{2}}{7}

The Conjugate Mirror

Now, turn your gaze to the green conjugate hyperbola,

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

. Here, the roles of

a

and

b

are effectively swapped. Its eccentricity squared is:

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

Its latus rectum is given by:

ℓ^{'} = \frac{2 a ^{2}}{b}

We are given the condition

(e^{'})^{2} = \frac{11}{8} ℓ^{'}

. Substituting these values, we find:

\frac{a ^{2} + b ^{2}}{b ^{2}} = \frac{11}{8} (\frac{2 a ^{2}}{b})

Simplifying this, we arrive at our second pillar:

\frac{a ^{2} + b ^{2}}{b} = \frac{11 a ^{2}}{4}

The Algebraic Dance

We now have two equations, both containing the term

a^{2} + b^{2}

. To solve for the relationship between

a

and

b

, we divide the first equation by the second:

\frac{\frac{a ^{2} + b ^{2}}{a}}{\frac{a ^{2} + b ^{2}}{b}} = \frac{\frac{11 b ^{2}}{7}}{\frac{11 a ^{2}}{4}}

The term

a^{2} + b^{2}

cancels out completely. We are left with:

\frac{b}{a} = \frac{4 b ^{2}}{7 a ^{2}}

Since

a

and

b

are positive, we divide by

\frac{b}{a}

to find the beautiful relationship:

1 = \frac{4 b}{7 a} \Rightarrow 7 a = 4 b

The Final Sprint

We substitute

b = \frac{7 a}{4}

back into our second equation:

\frac{a ^{2} + ( \frac{7 a}{4} ) ^{2}}{\frac{7 a}{4}} = \frac{11 a ^{2}}{4}

The numerator simplifies to

\frac{65 a ^{2}}{16}

. Dividing by

\frac{7 a}{4}

yields:

\frac{65 a}{28} = \frac{11 a ^{2}}{4}

Solving for

a

(assuming $a eq 0$), we find

a = \frac{65}{77}

. Consequently, we determine

b

b = \frac{7}{4} \times \frac{65}{77} = \frac{65}{44}

Finally, we calculate the requested value

77 a + 44 b

77 (\frac{65}{77}) + 44 (\frac{65}{44}) = 65 + 65 = 130

The final answer is 130.

Common Traps & FAQs

Wait, why do we reject the negative root or why don't we solve for

e

directly?

Great question! In JEE Advanced problems, solving for

e

directly often introduces unnecessary square roots and complicates the algebra. By keeping the equations in terms of

e^{2}

and

(e^{'})^{2}

, we work with rational expressions that are much easier to manipulate. We don't reject roots here because

a

and

b

are defined as positive lengths, so we are strictly in the first quadrant of the parameter space.

How do I know when to divide the two equations?

This is a classic 'JEE Trap Catcher' technique. Whenever you see two equations that share a common, complex, or non-linear term (like

a^{2} + b^{2}

here), dividing them is often the fastest way to eliminate that term. It turns a system of non-linear equations into a simple ratio, which is much easier to solve.

Is there a geometric interpretation of the final sum

77 a + 44 b

Not necessarily a deep geometric one! In many competitive exams, the final expression is designed to 'clean up' the fractions you derived. Notice how the denominators of our final values for

a

and

b

were

77

and

44

. The question asks for

77 a + 44 b

precisely to make those denominators cancel out, rewarding you for your algebraic precision.

Select Answer:

Visualized Solution

Introduction to Hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$

Properties of Hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$

Applying the First Condition $e^{2} = \frac{11}{14} ℓ$

Simplifying the First Equation

Properties of Conjugate Hyperbola $\frac{y ^{2}}{b ^{2}} - \frac{x ^{2}}{a ^{2}} = 1$

Applying the Second Condition $(e^{'})^{2} = \frac{11}{8} ℓ^{'}$

Simplifying the Second Equation

Dividing Equation (1) by (2)

Finding the Ratio of $a$ and $b$

Substituting $b = \frac{7 a}{4}$ into Equation (2)

Solving for $a$

Calculating the Value of $b$

Final Calculation: $77 a + 44 b$

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

The Anatomy of the Hyperbola

The Conjugate Mirror

The Algebraic Dance

The Final Sprint

Similar Questions

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 foci and length of the latus rectum of an ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, a > b$ be $(\pm 5, 0)$ and $50$ , respectively. Then, the square of the eccentricity of the hyperbola $\frac{x ^{2}}{b ^{2}} - \frac{y ^{2}}{a ^{2} b ^{2}} = 1$ equals

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 the length of a latus rectum of an ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ be 10 . If its eccentricity is the minimum value of the function $f (t) = t^{2} + t + \frac{11}{12}, t \in R$ , then $a^{2} + b^{2}$ is equal to :

The foci of the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{b ^{2}} = 1$ and the hyperbola $\frac{x ^{2}}{144} - \frac{y ^{2}}{81} = \frac{1}{25}$ coincide. Then the value of $b^{2}$ is

Let the foci of a hyperbola coincide with the foci of the ellipse $\frac{x ^{2}}{36} + \frac{y ^{2}}{16} = 1$ . If the eccentricity of the hyperbola is 5, then the length of its latus rectum is :

Select Answer:

Visualized Solution

Introduction to Hyperbola .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2x2​−b2y2​=1

Properties of Hyperbola .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2x2​−b2y2​=1

Applying the First Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e2=1411​ℓ

Simplifying the First Equation

Properties of Conjugate Hyperbola .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2y2​−a2x2​=1

Applying the Second Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (e′)2=811​ℓ′

Simplifying the Second Equation

Dividing Equation (1) by (2)

Finding the Ratio of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b=47a​ into Equation (2)

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

Calculating the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Final Calculation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 77a+44b

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

The Anatomy of the Hyperbola

The Conjugate Mirror

The Algebraic Dance

The Final Sprint

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The foci of the ellipse 16x2​+b2y2​=1 and the hyperbola 144x2​−81y2​=251​ coincide. Then the value of b2 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the foci of a hyperbola coincide with the foci of the ellipse 36x2​+16y2​=1. If the eccentricity of the hyperbola is 5, then the length of its latus rectum is :

Introduction to Hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$

Properties of Hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$

Applying the First Condition $e^{2} = \frac{11}{14} ℓ$

Properties of Conjugate Hyperbola $\frac{y ^{2}}{b ^{2}} - \frac{x ^{2}}{a ^{2}} = 1$

Applying the Second Condition $(e^{'})^{2} = \frac{11}{8} ℓ^{'}$

Finding the Ratio of $a$ and $b$

Substituting $b = \frac{7 a}{4}$ into Equation (2)

Solving for $a$

Calculating the Value of $b$

Final Calculation: $77 a + 44 b$

The foci of the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{b ^{2}} = 1$ and the hyperbola $\frac{x ^{2}}{144} - \frac{y ^{2}}{81} = \frac{1}{25}$ coincide. Then the value of $b^{2}$ is

Let the foci of a hyperbola coincide with the foci of the ellipse $\frac{x ^{2}}{36} + \frac{y ^{2}}{16} = 1$ . If the eccentricity of the hyperbola is 5, then the length of its latus rectum is :