JEE Main 2022 (24 June Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: Let the hyperbola $H : \frac{x ^{2}}{a ^{2}} - y^{2} = 1$ and the ellipse $E : 3 x^{2} + 4 y^{2} = 12$ be such that the length of latus rectum of $H$ is equal to the length of latus rectum of $E$ . If $e_{H}$ and $e_{E}$ are the eccentricities of $H$ and $E$ respectively, then the value of $12 (e_{H}^{2} + e_{E}^{2})$ is equal to ______.

Q: Wait, why is the hyperbola's $b^2$ equal to $1$? The equation is just $\frac{x^2}{a^2} - y^2 = 1$.

Great observation! In the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the term $y^2$ is mathematically equivalent to $\frac{y^2}{1}$. Thus, $b^2$ is indeed $1$. Always look for that hidden denominator in the standard form—it is a classic JEE trap!

Q: Is there a shortcut to finding the latus rectum?

While you can memorize $\frac{2b^2}{a}$, I encourage you to visualize it as the chord passing through the focus perpendicular to the axis. Deriving it once will make you immune to forgetting it during the exam! It is not just a formula; it is a geometric property of the conic.

Enter Numerical Value:

Visualized Solution

Standard Form of Ellipse $E$

Given Ellipse $E$ : $3 x^{2} + 4 y^{2} = 12$
Divide the entire equation by $12$ :
$\frac{3 x ^{2}}{12} + \frac{4 y ^{2}}{12} = 1$
Standard form: $\frac{x ^{2}}{4} + \frac{y ^{2}}{3} = 1$

Parameters of Ellipse $E$

Comparing with $\frac{x ^{2}}{a _{E}^{2}} + \frac{y ^{2}}{b _{E}^{2}} = 1$
$a_{E}^{2} = 4 ⟹ a_{E} = 2$
$b_{E}^{2} = 3$

Latus Rectum of Ellipse $E$

Formula for Length of Latus Rectum: $L R_{E} = \frac{2 b _{E}^{2}}{a _{E}}$
Substitute $b_{E}^{2} = 3$ and $a_{E} = 2$ :
$L R_{E} = \frac{2 ( 3 )}{2}$
$L R_{E} = 3$

Eccentricity of Ellipse $e_{E}$

Formula for Eccentricity: $e_{E}^{2} = 1 - \frac{b _{E}^{2}}{a _{E}^{2}}$
Substitute the values:
$e_{E}^{2} = 1 - \frac{3}{4}$
$e_{E}^{2} = \frac{1}{4}$

Standard Form of Hyperbola $H$

Given Hyperbola $H$ : $\frac{x ^{2}}{a ^{2}} - y^{2} = 1$
Rewrite in standard form:
$\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{1} = 1$
Here, $a_{H}^{2} = a^{2}$ and $b_{H}^{2} = 1$

Latus Rectum of Hyperbola $H$

Formula for Length of Latus Rectum: $L R_{H} = \frac{2 b _{H}^{2}}{a _{H}}$
Substitute $b_{H}^{2} = 1$ and $a_{H} = a$ :
$L R_{H} = \frac{2 ( 1 )}{a}$
$L R_{H} = \frac{2}{a}$

Equating the Latus Recta

Given condition: $L R_{H} = L R_{E}$
Substitute the calculated lengths:
$\frac{2}{a} = 3$
Solving for $a$ :
$a = \frac{2}{3}$

Eccentricity of Hyperbola $e_{H}$ Setup

Formula for Eccentricity: $e_{H}^{2} = 1 + \frac{b _{H}^{2}}{a _{H}^{2}}$
We know $b_{H}^{2} = 1$ and $a_{H}^{2} = a^{2}$
Substitute $a = \frac{2}{3} ⟹ a^{2} = \frac{4}{9}$ :
$e_{H}^{2} = 1 + \frac{1}{\frac{4}{9}}$

Computing $e_{H}^{2}$

Simplify the fraction:
$e_{H}^{2} = 1 + \frac{9}{4}$
Take the common denominator:
$e_{H}^{2} = \frac{4 + 9}{4} = \frac{13}{4}$

Setting Up the Final Expression

We need to find the value of: $12 (e_{H}^{2} + e_{E}^{2})$
Substitute $e_{H}^{2} = \frac{13}{4}$ and $e_{E}^{2} = \frac{1}{4}$ :
Expression = $12 (\frac{13}{4} + \frac{1}{4})$

Final Calculation

Add the fractions inside the bracket:
$\frac{13}{4} + \frac{1}{4} = \frac{14}{4}$
Multiply by $12$ :
$12 \times \frac{14}{4} = 3 \times 14$
Final Answer = $42$

00:00 / 00:00

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

Solution Diagram

Analyzing the Ellipse

We start with the ellipse,

E

, defined by the equation

3 x^{2} + 4 y^{2} = 12

. To find the standard form, we divide the entire equation by

12

\frac{3 x ^{2}}{12} + \frac{4 y ^{2}}{12} = 1 \Rightarrow \frac{x ^{2}}{4} + \frac{y ^{2}}{3} = 1

Comparing this to the standard form

\frac{x ^{2}}{a _{E}^{2}} + \frac{y ^{2}}{b _{E}^{2}} = 1

, we identify

a_{E}^{2} = 4

(so

a_{E} = 2

) and

b_{E}^{2} = 3

The length of the latus rectum,

L R_{E}

, is given by the formula

\frac{2 b _{E}^{2}}{a _{E}}

. Substituting our values:

L R_{E} = \frac{2 ( 3 )}{2} = 3

Next, we calculate the square of the eccentricity,

e_{E}^{2}

, using the formula

e_{E}^{2} = 1 - \frac{b _{E}^{2}}{a _{E}^{2}}

e_{E}^{2} = 1 - \frac{3}{4} = \frac{1}{4}

Analyzing the Hyperbola

Now, we consider the hyperbola,

H

, given by

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

. We rewrite this as

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

, where

a_{H}^{2} = a^{2}

and

b_{H}^{2} = 1

The length of the latus rectum for the hyperbola,

L R_{H}

, is

\frac{2 b _{H}^{2}}{a _{H}}

. Substituting our values:

L R_{H} = \frac{2 ( 1 )}{a} = \frac{2}{a}

The Bridge

Equating the Latus Recta

We are given that the length of the latus rectum of the hyperbola is equal to that of the ellipse. Setting

L R_{H} = L R_{E}

, we obtain:

\frac{2}{a} = 3 \Rightarrow a = \frac{2}{3}

With

a = \frac{2}{3}

, we find

a^{2} = \frac{4}{9}

. We now calculate the square of the eccentricity for the hyperbola,

e_{H}^{2}

, using

e_{H}^{2} = 1 + \frac{b _{H}^{2}}{a _{H}^{2}}

e_{H}^{2} = 1 + \frac{1}{\frac{4}{9}} = 1 + \frac{9}{4} = \frac{13}{4}

Final Calculation

The problem asks for the value of

12 (e_{H}^{2} + e_{E}^{2})

. Substituting our derived values for

e_{H}^{2}

and

e_{E}^{2}

12 (\frac{13}{4} + \frac{1}{4}) = 12 (\frac{14}{4})

Simplifying the expression:

12 \times \frac{14}{4} = 3 \times 14 = 42

The final result is 42.

Common Traps & FAQs

Wait, why is the hyperbola's

b^{2}

equal to

1

? The equation is just

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

Great observation! In the standard form

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

, the term

y^{2}

is mathematically equivalent to

\frac{y ^{2}}{1}

. Thus,

b^{2}

is indeed

1

. Always look for that hidden denominator in the standard form—it is a classic JEE trap!

Why do we use different formulas for eccentricity? Isn't it always

1 - \frac{b ^{2}}{a ^{2}}

That is a classic trap! For an ellipse, the curve is 'contained,' so

e < 1

, leading to the subtraction formula

1 - \frac{b ^{2}}{a ^{2}}

. For a hyperbola, the curve is 'open,' so

e > 1

, which requires the addition formula

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

. Think of the hyperbola as an 'expanded' ellipse where the focus is pushed further out.

Is there a shortcut to finding the latus rectum?

While you can memorize

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

, I encourage you to visualize it as the chord passing through the focus perpendicular to the axis. Deriving it once will make you immune to forgetting it during the exam! It is not just a formula; it is a geometric property of the conic.

Enter Numerical Value:

Visualized Solution

Standard Form of Ellipse $E$

Parameters of Ellipse $E$

Latus Rectum of Ellipse $E$

Eccentricity of Ellipse $e_{E}$

Standard Form of Hyperbola $H$

Latus Rectum of Hyperbola $H$

Equating the Latus Recta

Eccentricity of Hyperbola $e_{H}$ Setup

Computing $e_{H}^{2}$

Setting Up the Final Expression

Final Calculation

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

Analyzing the Ellipse

Analyzing the Hyperbola

The Bridge

Final Calculation

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 eccentricity of an ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ is reciprocal to that of the hyperbola $2 x^{2} - 2 y^{2} = 1$ . If the ellipse intersects the hyperbola at right angles, then square of length of the latus-rectum of the ellipse 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 :

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 $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, a > b$ 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 foci of the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{7} = 1$ and the hyperbola $\frac{x ^{2}}{144} - \frac{y ^{2}}{α} = \frac{1}{25}$ coincide. Then the length of the latus rectum of the hyperbola is:-

Enter Numerical Value:

Visualized Solution

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

Parameters of Ellipse .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E

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

Eccentricity of Ellipse .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } eE​

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

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

Equating the Latus Recta

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

Computing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } eH2​

Setting Up the Final Expression

Final Calculation

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

Analyzing the Ellipse

Analyzing the Hyperbola

The Bridge

Final Calculation

Similar Questions

.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 :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the foci of the ellipse 16x2​+7y2​=1 and the hyperbola 144x2​−αy2​=251​ coincide. Then the length of the latus rectum of the hyperbola is:-

Standard Form of Ellipse $E$

Parameters of Ellipse $E$

Latus Rectum of Ellipse $E$

Eccentricity of Ellipse $e_{E}$

Standard Form of Hyperbola $H$

Latus Rectum of Hyperbola $H$

Eccentricity of Hyperbola $e_{H}$ Setup

Computing $e_{H}^{2}$

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 :

Let the foci of the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{7} = 1$ and the hyperbola $\frac{x ^{2}}{144} - \frac{y ^{2}}{α} = \frac{1}{25}$ coincide. Then the length of the latus rectum of the hyperbola is:-