JEE Main 2026 (28 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: 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 :

Q: Does the latus rectum formula change for a vertical hyperbola?

The formula $L = \frac{2B^2}{A}$ remains the same, but you must be careful with your definitions of $A$ and $B$. $A$ is always the semi-axis associated with the positive term (the transverse axis), and $B$ is the semi-axis associated with the negative term (the conjugate axis). Always identify the transverse axis first!

Q: Why do we need to find $\lambda$ if we only need $e$ and $L$?

You need $\lambda$ because it defines the semi-transverse axis $A$. Without knowing the specific dimensions of the hyperbola, you cannot determine the eccentricity or the latus rectum. Think of $\lambda$ as the key that unlocks the geometry of the hyperbola.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark: Look closely at the denominators of the ellipse equation. Since the larger denominator is under the

y^{2}

term, this is a vertical ellipse. The foci will lie on the y-axis. Do you remember how the focal distance

c

relates to the semi-major and semi-minor axes for an ellipse?

Hint 2

Hint 2: The Arsenal: For a hyperbola, the standard form is

\frac{y ^{2}}{A ^{2}} - \frac{x ^{2}}{B ^{2}} = 1

when it is vertical. Recall that for a hyperbola, the relationship between the focal distance

c

, the semi-transverse axis

A

, and the semi-conjugate axis

B

c^{2} = A^{2} + B^{2}

. Use the fact that the foci are identical to bridge the two curves.

Hint 3

Hint 3: The Execution: Once you have identified

c^{2} = 25

from the ellipse, set it equal to the hyperbola's

A^{2} + B^{2}

. You know

B^{2} = 16

from the hyperbola's equation. Solve for

A^{2}

, then calculate

e = \frac{c}{A}

and

L = \frac{2 B ^{2}}{A}

to reach your final destination.

Visualized Solution

Analyze Ellipse $E$ Orientation

Ellipse $E : \frac{x ^{2}}{144} + \frac{y ^{2}}{169} = 1$
Comparing with $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$
Since $169 > 144$ , it is a vertical ellipse ( $b > a$ )

Focal Distance Formula for Ellipse

For a vertical ellipse, $c^{2} = b^{2} - a^{2}$
Here, $b^{2} = 169$ and $a^{2} = 144$

Calculate Foci of Ellipse

$c^{2} = 169 - 144 = 25$
Therefore, $c = 5$
Foci lie on the y-axis: $(0, \pm c) ⟹ (0, \pm 5)$

Standardize Hyperbola Equation

Hyperbola $H : \frac{x ^{2}}{16} - \frac{y ^{2}}{λ ^{2}} = - 1$
Multiply by $- 1$ : $\frac{y ^{2}}{λ ^{2}} - \frac{x ^{2}}{16} = 1$
This is a vertical hyperbola

Focal Relation for Hyperbola

For vertical hyperbola, $c_{H}^{2} = A^{2} + B^{2}$
Here, $A^{2} = λ^{2}$ and $B^{2} = 16$

Equate Foci of Ellipse and Hyperbola

Given: $E$ and $H$ have the same foci
Therefore, $c_{H}^{2} = c^{2} = 25$
Substitute: $λ^{2} + 16 = 25$

Solve for $λ^{2}$

$λ^{2} = 25 - 16 = 9$
Therefore, $λ = 3$
Hyperbola parameters: $A = 3$ , $B = 4$

Calculate Eccentricity $e$

Eccentricity $e = \frac{c}{A}$
Substitute $c = 5$ and $A = 3$
$e = \frac{5}{3}$

Calculate Latus Rectum $L$

Length of Latus Rectum $L = \frac{2 B ^{2}}{A}$
Substitute $B^{2} = 16$ and $A = 3$
$L = \frac{2 ( 16 )}{3} = \frac{32}{3}$

Setup Final Expression

We need to find the value of $24 (e + L)$
Substitute $e = \frac{5}{3}$ and $L = \frac{32}{3}$
Expression: $24 (\frac{5}{3} + \frac{32}{3})$

Final Computation

$e + L = \frac{5 + 32}{3} = \frac{37}{3}$
$24 \times \frac{37}{3} = 8 \times 37$
Final Answer: $296$

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

Solution Diagram

Analyzing the Ellipse

We begin by examining the equation of the ellipse:

\frac{x ^{2}}{144} + \frac{y ^{2}}{169} = 1

Since

169 > 144

, the ellipse is stretched along the vertical axis. We identify the semi-major axis

b = 169 = 13

and the semi-minor axis

a = 144 = 12

Unveiling the Hidden Foci

To locate the foci, we utilize the relationship

c^{2} = b^{2} - a^{2}

. This distance

c

represents the displacement of the foci from the center of the ellipse.

Substituting our known values:

c^{2} = 169 - 144 = 25

This yields

c = 5

. Consequently, the foci of the ellipse are located at

(0, 5)

and

(0, - 5)

The Hyperbola's Transformation

Next, we consider the hyperbola

H

defined by:

\frac{x ^{2}}{16} - \frac{y ^{2}}{λ ^{2}} = - 1

By multiplying the entire equation by

- 1

, we reveal its standard form:

\frac{y ^{2}}{λ ^{2}} - \frac{x ^{2}}{16} = 1

This confirms the hyperbola is vertical. Here, the semi-transverse axis is

A = λ

and the semi-conjugate axis is

B = 4

The Bridge of Foci

The problem states that the ellipse and the hyperbola share the same foci. For a hyperbola, the focal distance

c_{H}

is governed by the relation

c_{H}^{2} = A^{2} + B^{2}

Setting

c_{H}^{2}

equal to the ellipse's

c^{2} = 25

, we obtain:

λ^{2} + 16 = 25

Solving for

λ^{2}

, we find

λ^{2} = 9

, which implies

λ = 3

. Thus, for our hyperbola,

A = 3

and

B = 4

The Final Synthesis

With the parameters defined, we calculate the eccentricity

e

and the length of the latus rectum

L

e = \frac{c}{A} = \frac{5}{3}

L = \frac{2 B ^{2}}{A} = \frac{2 ( 16 )}{3} = \frac{32}{3}

We are tasked with finding the value of

24 (e + L)

. Substituting our results:

24 (\frac{5}{3} + \frac{32}{3}) = 24 (\frac{37}{3})

Performing the final arithmetic,

8 \times 37 = 296

. The final result is 296.

Common Traps & FAQs

Wait, why is the hyperbola vertical? The equation looks like

\frac{x ^{2}}{16} - \frac{y ^{2}}{λ ^{2}} = - 1

Great catch! When you see a negative sign on the right side, it is a massive red flag. Always multiply the entire equation by

- 1

to standardize it. You get

\frac{y ^{2}}{λ ^{2}} - \frac{x ^{2}}{16} = 1

. Now, the positive term is under

y^{2}

, which confirms it is a vertical hyperbola opening along the y-axis.

Why did we use

c^{2} = b^{2} - a^{2}

for the ellipse but

c^{2} = A^{2} + B^{2}

for the hyperbola?

Think of the geometry! In an ellipse, the foci are inside the curve, so the distance

c

must be smaller than the semi-major axis

b

. Thus,

c^{2} = b^{2} - a^{2}

. In a hyperbola, the foci are 'outside' the vertices, making

c

the longest segment in that right triangle relationship, hence

c^{2} = A^{2} + B^{2}

. It is all about where the points sit relative to the center.

Is the eccentricity

e

always greater than 1 for a hyperbola?

Absolutely! That is the defining characteristic of a hyperbola. If you ever calculate an

e < 1

for a hyperbola, you know you have made a calculation error. Here,

e = \frac{5}{3} \approx 1.66

, which perfectly fits the definition.

Does the latus rectum formula change for a vertical hyperbola?

The formula

L = \frac{2 B ^{2}}{A}

remains the same, but you must be careful with your definitions of

A

and

B

A

is always the semi-axis associated with the positive term (the transverse axis), and

B

is the semi-axis associated with the negative term (the conjugate axis). Always identify the transverse axis first!

Why do we need to find

λ

if we only need

e

and

L

You need

λ

because it defines the semi-transverse axis

A

. Without knowing the specific dimensions of the hyperbola, you cannot determine the eccentricity or the latus rectum. Think of

λ

as the key that unlocks the geometry of the hyperbola.

Select Answer:

Visualized Solution

Analyze Ellipse $E$ Orientation

Focal Distance Formula for Ellipse

Calculate Foci of Ellipse

Standardize Hyperbola Equation

Focal Relation for Hyperbola

Equate Foci of Ellipse and Hyperbola

Solve for $λ^{2}$

Calculate Eccentricity $e$

Calculate Latus Rectum $L$

Setup Final Expression

Final Computation

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

Analyzing the Ellipse

Unveiling the Hidden Foci

The Hyperbola's Transformation

The Bridge of Foci

The Final Synthesis

Similar Questions

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:-

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 a hyperbola be $(1, 14)$ and $(1, - 12)$ . If it passes through the point $(1, 6),$ 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 :

Select Answer:

Visualized Solution

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

Focal Distance Formula for Ellipse

Calculate Foci of Ellipse

Standardize Hyperbola Equation

Focal Relation for Hyperbola

Equate Foci of Ellipse and Hyperbola

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ2

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

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

Setup Final Expression

Final Computation

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

Analyzing the Ellipse

Unveiling the Hidden Foci

The Hyperbola's Transformation

The Bridge of Foci

The Final Synthesis

Similar Questions

.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:-

.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 a hyperbola be (1,14) and (1,−12). If it passes through the point (1,6), then the length of its latus-rectum is :

Analyze Ellipse $E$ Orientation

Solve for $λ^{2}$

Calculate Eccentricity $e$

Calculate Latus Rectum $L$

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:-

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 a hyperbola be $(1, 14)$ and $(1, - 12)$ . If it passes through the point $(1, 6),$ then the length of its latus-rectum is :