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 $\sqrt{50}$, respectively. Then, the square of the eccentricity of the hyperbola $\frac{x^2}{b^2} - \frac{y^2}{a^2 b^2} = 1$ equals

### Analyzing the Ellipse's Blueprint Imagine you are standing on the Cartesian plane. We have an ellipse, $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ with $a > b$. This tells us immediately that our ellipse is stretched horizontally. The problem provides two vital clues: the foci at $(\pm 5, 0)$ and a latus rectum length of $\sqrt{50}$. Let us start with the foci. In any standard ellipse, the foci are located at $(\pm ae, 0)$. By looking at our given coordinates $(\pm 5, 0)$, we deduce that $ae = 5$. This is our first anchor. Next, consider the latus rectum. This is the chord passing through the focus, perpendicular to the major axis. Its length is defined by the formula: $$LR = \frac{2b^2}{a}$$ We are told this length is $\sqrt{50}$, which simplifies to $5\sqrt{2}$. Thus, we have our second equation: $$\frac{2b^2}{a} = 5\sqrt{2}$$ ### The Algebraic Bridge We recall the fundamental identity of an ellipse: $b^2 = a^2(1 - e^2)$. If we distribute the $a^2$, we get $b^2 = a^2 - (ae)^2$. Since we know $ae = 5$, it follows that $(ae)^2 = 25$. Thus, our identity becomes $b^2 = a^2 - 25$. We now have two expressions for $b^2$: 1. From the latus rectum: $b^2 = \frac{5\sqrt{2}a}{2}$ 2. From the fundamental identity: $b^2 = a^2 - 25$ Since both equal $b^2$, we set them equal: $a^2 - 25 = \frac{5\sqrt{2}a}{2}$. Clearing the fraction by multiplying by $2$, we obtain $2a^2 - 50 = 5\sqrt{2}a$. Rearranging this into a standard quadratic form, we get: $$2a^2 - 5\sqrt{2}a - 50 = 0$$ Using the quadratic formula $a = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$, we find the roots. The discriminant is $(5\sqrt{2})^2 - 4(2)(-50) = 50 + 400 = 450$. The square root of $450$ is $15\sqrt{2}$. Thus, $a = \frac{5\sqrt{2} \pm 15\sqrt{2}}{4}$. Since $a$ must be positive, we take the positive root: $a = \frac{20\sqrt{2}}{4} = 5\sqrt{2}$. Squaring this, we find **$a^2 = 50$**, and consequently, **$b^2 = 25$**. ### The Hyperbola's Transformation We are given a hyperbola: $\frac{x^2}{b^2} - \frac{y^2}{a^2b^2} = 1$. Let us map this to the standard hyperbola form $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$, where $A^2 = b^2$ and $B^2 = a^2b^2$. The question asks for the square of the eccentricity, $e_H^2$. The formula for the eccentricity of a hyperbola is $e_H^2 = 1 + \frac{B^2}{A^2}$. Substituting our values, we get: $$e_H^2 = 1 + \frac{a^2b^2}{b^2}$$ The $b^2$ terms cancel out perfectly, leaving us with $e_H^2 = 1 + a^2$. Since we already determined $a^2 = 50$, we calculate: $$e_H^2 = 1 + 50 = 51$$ ### Final Result Through the systematic application of geometric identities and algebraic manipulation, we have arrived at the final value. The square of the eccentricity of the hyperbola is **$51$**.

JEE Main 2024 (31 Jan Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: 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

Enter Numerical Value:

Progressive Hints

Hint 1

The Spark: Visualize the ellipse as a geometric system defined by its constraints. You are given the foci

(\pm 5, 0)

and the latus rectum length

50

. Before diving into algebra, realize that the foci tell you the value of

a e

, and the latus rectum gives you a direct relationship between

a

and

b

Hint 2

The Arsenal: You need two primary tools here. First, the fundamental ellipse identity

b^{2} = a^{2} (1 - e^{2})

, which simplifies to

b^{2} = a^{2} - (a e)^{2}

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

L R = \frac{2 b ^{2}}{a}

. Use these to create a system of equations.

Hint 3

The Execution: Once you have the quadratic equation in terms of

a

, solve for

a^{2}

. Do not worry about finding

a

and

b

individually if you don't have to. For the hyperbola, identify its semi-axes

A^{2} = b^{2}

and

B^{2} = a^{2} b^{2}

, then apply the eccentricity formula

e_{H}^{2} = 1 + \frac{B ^{2}}{A ^{2}}

Visualized Solution

Visualizing the Ellipse

Given Ellipse: $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ where $a > b$ .
Foci are located at $(\pm 5, 0)$ .

Foci Relationship

Standard Foci coordinates for $a > b$ : $(\pm a e, 0)$ .
Comparing with given foci: $a e = 5$ .

Latus Rectum

Length of Latus Rectum ( $L R$ ) is given as $50$ .
$50$ simplifies to $52$ .

Latus Rectum Formula

Formula for $L R$ : $\frac{2 b ^{2}}{a}$ .
Equating to given length: $\frac{2 b ^{2}}{a} = 52$ .
Rearranging for $b^{2}$ : $b^{2} = \frac{5 2 a}{2}$ .

The Ellipse Identity

Fundamental Relation for Ellipse: $b^{2} = a^{2} (1 - e^{2})$ .
Expanding the terms: $b^{2} = a^{2} - a^{2} e^{2}$ .
This can be written as: $b^{2} = a^{2} - (a e)^{2}$ .

Substituting $a e$

Recall from earlier: $a e = 5$ .
Substitute into the identity: $b^{2} = a^{2} - (5)^{2}$ .
Simplifying: $b^{2} = a^{2} - 25$ .

Equating Expressions for $b^{2}$

We have two expressions for $b^{2}$ :
1. $b^{2} = \frac{5 2 a}{2}$
2. $b^{2} = a^{2} - 25$
Equating them: $a^{2} - 25 = \frac{5 2 a}{2}$ .

Forming the Quadratic Equation

Multiply the entire equation by $2$ :
$2 (a^{2} - 25) = 52 a$
Expand: $2 a^{2} - 50 = 52 a$ .
Rearrange into standard quadratic form:
$2 a^{2} - 52 a - 50 = 0$ .

Solving for $a$

Use the quadratic formula: $a = \frac{- B \pm B ^{2} - 4 A C}{2 A}$ .
Substitute $A = 2, B = - 52, C = - 50$ :
$a = \frac{5 2 \pm ( - 5 2 ) ^{2} - 4 ( 2 ) ( - 50 )}{2 ( 2 )}$ .
Simplify discriminant: $50 + 400 = 450$ .
$a = \frac{5 2 \pm 15 2}{4}$ .

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

Since $a > 0$ , take the positive root:
$a = \frac{20 2}{4} = 52$ .
Calculate $a^{2}$ : $a^{2} = (52)^{2} = 50$ .
Calculate $b^{2}$ : $b^{2} = a^{2} - 25 = 50 - 25 = 25$ .

Introducing the Hyperbola

Given Hyperbola: $\frac{x ^{2}}{b ^{2}} - \frac{y ^{2}}{a ^{2} b ^{2}} = 1$ .
Let's write it in standard form: $\frac{x ^{2}}{A ^{2}} - \frac{y ^{2}}{B ^{2}} = 1$ .
Here, $A^{2} = b^{2}$ and $B^{2} = a^{2} b^{2}$ .

Hyperbola Eccentricity Formula

We need the square of the eccentricity of the hyperbola, let's call it $e_{H}^{2}$ .
Formula: $e_{H}^{2} = 1 + \frac{B ^{2}}{A ^{2}}$ .

Substituting Hyperbola Parameters

Substitute $A^{2} = b^{2}$ and $B^{2} = a^{2} b^{2}$ into the formula:
$e_{H}^{2} = 1 + \frac{a ^{2} b ^{2}}{b ^{2}}$ .
Notice that $b^{2}$ cancels out from the numerator and denominator.
Simplified relation: $e_{H}^{2} = 1 + a^{2}$ .

Final Calculation

We already found $a^{2} = 50$ .
Substitute this into our simplified relation:
$e_{H}^{2} = 1 + 50$ .
Final Answer: $e_{H}^{2} = 51$ .

00:00 / 00:00

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

Solution Diagram

Analyzing the Ellipse's Blueprint

Imagine you are standing on the Cartesian plane. We have an ellipse,

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

with

a > b

. This tells us immediately that our ellipse is stretched horizontally. The problem provides two vital clues: the foci at

(\pm 5, 0)

and a latus rectum length of

50

Let us start with the foci. In any standard ellipse, the foci are located at

(\pm a e, 0)

. By looking at our given coordinates

(\pm 5, 0)

, we deduce that

a e = 5

. This is our first anchor.

Next, consider the latus rectum. This is the chord passing through the focus, perpendicular to the major axis. Its length is defined by the formula:

L R = \frac{2 b ^{2}}{a}

We are told this length is

50

, which simplifies to

52

. Thus, we have our second equation:

\frac{2 b ^{2}}{a} = 52

The Algebraic Bridge

We recall the fundamental identity of an ellipse:

b^{2} = a^{2} (1 - e^{2})

. If we distribute the

a^{2}

, we get

b^{2} = a^{2} - (a e)^{2}

. Since we know

a e = 5

, it follows that

(a e)^{2} = 25

. Thus, our identity becomes

b^{2} = a^{2} - 25

We now have two expressions for

b^{2}

: 1. From the latus rectum:

b^{2} = \frac{5 2 a}{2}

2. From the fundamental identity:

b^{2} = a^{2} - 25

Since both equal

b^{2}

, we set them equal:

a^{2} - 25 = \frac{5 2 a}{2}

. Clearing the fraction by multiplying by

2

, we obtain

2 a^{2} - 50 = 52 a

. Rearranging this into a standard quadratic form, we get:

2 a^{2} - 52 a - 50 = 0

Using the quadratic formula

a = \frac{- B \pm B ^{2} - 4 A C}{2 A}

, we find the roots. The discriminant is

(52)^{2} - 4 (2) (- 50) = 50 + 400 = 450

. The square root of

450

152

Thus,

a = \frac{5 2 \pm 15 2}{4}

. Since

a

must be positive, we take the positive root:

a = \frac{20 2}{4} = 52

. Squaring this, we find $a^{2} = 50$ , and consequently, $b^{2} = 25$ .

The Hyperbola's Transformation

We are given a hyperbola:

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

. Let us map this to the standard hyperbola form

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

, where

A^{2} = b^{2}

and

B^{2} = a^{2} b^{2}

The question asks for the square of the eccentricity,

e_{H}^{2}

. The formula for the eccentricity of a hyperbola is

e_{H}^{2} = 1 + \frac{B ^{2}}{A ^{2}}

. Substituting our values, we get:

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

The

b^{2}

terms cancel out perfectly, leaving us with

e_{H}^{2} = 1 + a^{2}

. Since we already determined

a^{2} = 50

, we calculate:

e_{H}^{2} = 1 + 50 = 51

Final Result

Through the systematic application of geometric identities and algebraic manipulation, we have arrived at the final value. The square of the eccentricity of the hyperbola is $51$ .

Common Traps & FAQs

Wait, why do we reject the negative root when solving the quadratic equation for

a

That is a fantastic question! In the context of conic sections,

a

represents the semi-major axis, which is a physical length. By definition, a length must be a positive scalar quantity. When we solve the quadratic equation

2 a^{2} - 52 a - 50 = 0

, we get two roots. The negative root would imply a negative length, which is physically impossible for the geometry of an ellipse. Always keep your physical intuition grounded in the math!

I got confused with the hyperbola equation. Why is the denominator

a^{2} b^{2}

It is easy to get tripped up by the notation! The problem defines the hyperbola as

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

. Here, the denominators are just labels for the square of the semi-axes. We simply map the standard form

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

to your equation. So,

A^{2} = b^{2}

and

B^{2} = a^{2} b^{2}

. Don't let the variables

a

and

b

from the ellipse distract you—treat them as constants once you have solved the first part of the problem.

Why does

b^{2}

cancel out in the hyperbola eccentricity calculation? Is that a coincidence?

It is not a coincidence at all—it is the beauty of mathematical design! When you calculate the eccentricity of the hyperbola, you use the ratio

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

. Since

B^{2} = a^{2} b^{2}

and

A^{2} = b^{2}

, the

b^{2}

term appears in both the numerator and the denominator. This cancellation is a hint that the eccentricity of this specific hyperbola depends solely on the parameter

a

of the original ellipse. Trust the algebra; it often simplifies to reveal the underlying structure.

Enter Numerical Value:

Visualized Solution

Visualizing the Ellipse

Foci Relationship

Latus Rectum

Latus Rectum Formula

The Ellipse Identity

Substituting $a e$

Equating Expressions for $b^{2}$

Forming the Quadratic Equation

Solving for $a$

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

Introducing the Hyperbola

Hyperbola Eccentricity Formula

Substituting Hyperbola Parameters

Final Calculation

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

Analyzing the Ellipse's Blueprint

The Algebraic Bridge

The Hyperbola's Transformation

Final Result

Similar Questions

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:-

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 _____.

Enter Numerical Value:

Visualized Solution

Visualizing the Ellipse

Foci Relationship

Latus Rectum

Latus Rectum Formula

The Ellipse Identity

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

Equating Expressions for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2

Forming the Quadratic Equation

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

Calculating .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

Introducing the Hyperbola

Hyperbola Eccentricity Formula

Substituting Hyperbola Parameters

Final Calculation

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

Analyzing the Ellipse's Blueprint

The Algebraic Bridge

The Hyperbola's Transformation

Final Result

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:-

Substituting $a e$

Equating Expressions for $b^{2}$

Solving for $a$

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

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:-