JEE Main 2026 (21 January Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: 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 :

Q: Can I just use the formula $e = \sqrt{1 + \frac{B^2}{A^2}}$ instead of $C = Ae$?

Absolutely! That is a perfectly valid approach. Since $e = \frac{C}{A}$, squaring both sides gives $e^2 = \frac{C^2}{A^2} = \frac{A^2 + B^2}{A^2} = 1 + \frac{B^2}{A^2}$. Both methods are mathematically equivalent; use whichever feels more intuitive to you.

Q: Why do we need $B^2$ instead of just $B$ in the final formula?

Great observation. The formula for the latus rectum is $LR = \frac{2B^2}{A}$. Since the calculation of $B^2$ comes directly from the identity $B^2 = C^2 - A^2$, it is much faster to work with $B^2$ directly rather than taking the square root to find $B$ and then squaring it again. Efficiency is key in JEE!

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. Visualize the ellipse and the hyperbola as two distinct geometric entities that share a common secret: their foci. The ellipse's equation

\frac{x ^{2}}{36} + \frac{y ^{2}}{16} = 1

allows you to determine its foci using the relationship

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

. Once you have these coordinates, you have unlocked the focal distance

C

for the hyperbola.

Hint 2

Hint 2: The Arsenal. Remember that for a hyperbola, the eccentricity

e

is defined as the ratio of the distance from the center to the focus (

C

) to the semi-transverse axis (

A

), expressed as

e = \frac{C}{A}

. You already have

e = 5

and the value of

C

from the ellipse. Use this to find

A

Hint 3

Hint 3: The Execution. Now that you have

A

and

C

, use the fundamental hyperbola relationship

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

to solve for

B^{2}

. Finally, apply the standard formula for the length of the latus rectum, which is

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

, to arrive at the final answer.

Visualized Solution

Analyze the Ellipse Equation

Given Ellipse: $\frac{x ^{2}}{36} + \frac{y ^{2}}{16} = 1$
Standard form: $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$
Comparing both, we get $a^{2} = 36$ and $b^{2} = 16$

Formula for Ellipse Foci

Distance of focus from center is $c$
Formula: $c = a^{2} - b^{2}$

Calculate Ellipse Foci

$c = 36 - 16 = 20$
$c = 25$
Foci of ellipse: $(\pm 25, 0)$

Transition to Hyperbola

Hyperbola shares the same foci.
For Hyperbola: $C = 25$
Eccentricity of hyperbola ( $e$ ) = $5$

Find Hyperbola's Semi-major Axis

Relation for eccentricity: $e = \frac{C}{A}$
Substitute $e = 5$ and $C = 25$
$5 = \frac{2 5}{A}$

Calculate $A$

Solving for $A$ :
$A = \frac{2 5}{5}$
$A = \frac{2}{5}$

Find $B^{2}$ for Hyperbola

Hyperbola relation: $C^{2} = A^{2} + B^{2}$
Rearranging: $B^{2} = C^{2} - A^{2}$

Calculate $B^{2}$

$B^{2} = (25)^{2} - (\frac{2}{5})^{2}$
$B^{2} = 20 - \frac{4}{5}$
$B^{2} = \frac{96}{5}$

Latus Rectum Formula

Length of Latus Rectum (LR) = $\frac{2 B ^{2}}{A}$

Substitute and Simplify LR

LR = $\frac{2 \cdot ( \frac{96}{5} )}{\frac{2}{5}}$
LR = $\frac{96}{5} \cdot 5$

Final Conclusion

LR = $\frac{96}{5 \cdot 5} \cdot 5$
LR = $\frac{96}{5}$
Final Answer: $\frac{96}{5}$

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

Solution Diagram

Analyzing the Setup

We begin our journey with the ellipse defined by the equation:

\frac{x ^{2}}{36} + \frac{y ^{2}}{16} = 1

This follows the standard form

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

, where

a^{2} = 36

and

b^{2} = 16

The distance of the focus from the center, denoted as

c

, is the key to our derivation. We use the fundamental relationship

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

Substituting our values, we find

c = 36 - 16 = 20 = 25

. Thus, the foci of our ellipse are located at

(\pm 25, 0)

The Bridge to the Hyperbola

The problem states that the hyperbola shares these exact same foci. For the hyperbola, the distance from the center to the focus is denoted by

C

Since the foci are identical, we immediately know

C = 25

. We are also given the eccentricity

e = 5

In the realm of hyperbolas, the eccentricity is defined as the ratio of the distance to the focus to the semi-transverse axis, expressed as

e = \frac{C}{A}

The Anatomy of the Hyperbola

With

e = 5

and

C = 25

, we can isolate the semi-transverse axis

A

5 = \frac{2 5}{A} \Rightarrow A = \frac{2 5}{5} = \frac{2}{5}

Next, we determine the semi-conjugate axis squared,

B^{2}

. We invoke the hyperbola's defining relationship:

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

Rearranging for

B^{2}

, we get:

B^{2} = C^{2} - A^{2} = (25)^{2} - (\frac{2}{5})^{2}

B^{2} = 20 - \frac{4}{5} = \frac{100 - 4}{5} = \frac{96}{5}

Final Calculation

The length of the latus rectum (

L R

) of a hyperbola is given by the formula:

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

Plugging in our derived values, we obtain:

L R = \frac{2 \cdot ( \frac{96}{5} )}{\frac{2}{5}}

The

2

in the numerator and the

2

in the denominator cancel out, leaving:

L R = \frac{96}{5} \cdot 5 = \frac{96}{5}

The final length of the latus rectum is $\frac{96}{5}$ .

Common Traps & FAQs

Wait, why is the formula for

c

different for the ellipse and the hyperbola? I always get confused!

That is a classic point of confusion! For an ellipse, the focus is inside the curve, so

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

. For a hyperbola, the focus is outside the vertices, meaning the distance to the focus must be larger than the semi-transverse axis, hence

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

. Think of it as the hyperbola being 'stretched' outward.

Can I just use the formula

e = 1 + \frac{B ^{2}}{A ^{2}}

instead of

C = A e

Absolutely! That is a perfectly valid approach. Since

e = \frac{C}{A}

, squaring both sides gives

e^{2} = \frac{C ^{2}}{A ^{2}} = \frac{A ^{2} + B ^{2}}{A ^{2}} = 1 + \frac{B ^{2}}{A ^{2}}

. Both methods are mathematically equivalent; use whichever feels more intuitive to you.

Why do we need

B^{2}

instead of just

B

in the final formula?

Great observation. The formula for the latus rectum is

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

. Since the calculation of

B^{2}

comes directly from the identity

B^{2} = C^{2} - A^{2}

, it is much faster to work with

B^{2}

directly rather than taking the square root to find

B

and then squaring it again. Efficiency is key in JEE!

Animated Solution for Mathematics - Conic Sections: 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

Analyze the Ellipse Equation

Formula for Ellipse Foci

Calculate Ellipse Foci

Transition to Hyperbola

Find Hyperbola's Semi-major Axis

Calculate $A$

Find $B^{2}$ for Hyperbola

Calculate $B^{2}$

Latus Rectum Formula

Substitute and Simplify LR

Final Conclusion

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

Analyzing the Setup

The Bridge to the Hyperbola

The Anatomy of the Hyperbola

Final Calculation

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:-

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 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 :

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 :

An ellipse has its center at $(1, - 2)$ , one focus at $(3, - 2)$ and one vertex at $(5, - 2)$ . Then the length of its latus rectum 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 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 _____.

The length of the latus-rectum of the ellipse, whose foci are $(2, 5)$ and $(2, - 3)$ and eccentricity is $\frac{4}{5}$ , is

Select Answer:

Visualized Solution

Analyze the Ellipse Equation

Formula for Ellipse Foci

Calculate Ellipse Foci

Transition to Hyperbola

Find Hyperbola's Semi-major Axis

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

Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B2 for Hyperbola

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B2

Latus Rectum Formula

Substitute and Simplify LR

Final Conclusion

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

Analyzing the Setup

The Bridge to the Hyperbola

The Anatomy of the Hyperbola

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 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 be (1,14) and (1,−12). If it passes through the point (1,6), 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; } An ellipse has its center at (1,−2), one focus at (3,−2) and one vertex at (5,−2). 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; } The length of the latus-rectum of the ellipse, whose foci are (2,5) and (2,−3) and eccentricity is 54​, is

Calculate $A$

Find $B^{2}$ for Hyperbola

Calculate $B^{2}$

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

An ellipse has its center at $(1, - 2)$ , one focus at $(3, - 2)$ and one vertex at $(5, - 2)$ . Then the length of its latus rectum is :

The length of the latus-rectum of the ellipse, whose foci are $(2, 5)$ and $(2, - 3)$ and eccentricity is $\frac{4}{5}$ , is