For some $\theta \in (0, \frac{\pi}{2})$, if the eccentricity of the hyperbola, $x^2 - y^2 \sec^2 \theta = 10$ is $\sqrt{5}$ times the eccentricity of the ellipse, $x^2 \sec^2 \theta + y^2 = 5$, then the length of the latus rectum of the ellipse, is : Options: (A) $2\sqrt{6}$ (B) $\frac{2\sqrt{5}}{3}$ (C) $\frac{4\sqrt{5}}{3}$ (D) $\sqrt{30}$

### Analyzing the Hyperbola The given equation for the hyperbola is $x^2 - y^2 \sec^2 \theta = 10$. Dividing by $10$, we obtain: $$\frac{x^2}{10} - \frac{y^2 \sec^2 \theta}{10} = 1$$ Using the identity $\frac{1}{\sec^2 \theta} = \cos^2 \theta$, the equation simplifies to: $$\frac{x^2}{10} - \frac{y^2}{10 \cos^2 \theta} = 1$$ The eccentricity of a hyperbola is defined by $e_H = \sqrt{1 + \frac{b^2}{a^2}}$. With $a^2 = 10$ and $b^2 = 10 \cos^2 \theta$, we find: $$e_H = \sqrt{1 + \frac{10 \cos^2 \theta}{10}} = \sqrt{1 + \cos^2 \theta}$$ ### Analyzing the Ellipse The equation for the ellipse is $x^2 \sec^2 \theta + y^2 = 5$. Dividing by $5$, we get: $$\frac{x^2 \sec^2 \theta}{5} + \frac{y^2}{5} = 1 \Rightarrow \frac{x^2}{5 \cos^2 \theta} + \frac{y^2}{5} = 1$$ Since $\theta \in (0, \frac{\pi}{2})$, we know $\cos^2 \theta < 1$, which implies $5 \cos^2 \theta < 5$. Because the denominator under $y^2$ is larger, the **major axis is vertical**. For a vertical ellipse, the eccentricity is $e_E = \sqrt{1 - \frac{a^2}{b^2}}$, where $a^2$ is the smaller denominator. Thus: $$e_E = \sqrt{1 - \frac{5 \cos^2 \theta}{5}} = \sqrt{1 - \cos^2 \theta} = \sin \theta$$ ### Solving for the Parameter We are given the relationship $e_H = \sqrt{5} e_E$. Substituting our expressions, we get: $$\sqrt{1 + \cos^2 \theta} = \sqrt{5} \sin \theta$$ Squaring both sides yields $1 + \cos^2 \theta = 5 \sin^2 \theta$. Using the identity $\sin^2 \theta = 1 - \cos^2 \theta$: $$1 + \cos^2 \theta = 5(1 - \cos^2 \theta)$$ $$1 + \cos^2 \theta = 5 - 5 \cos^2 \theta \Rightarrow 6 \cos^2 \theta = 4 \Rightarrow \cos^2 \theta = \frac{2}{3}$$ ### Final Calculation For our vertical ellipse, the semi-minor axis squared is $a^2 = 5 \cos^2 \theta = 5(\frac{2}{3}) = \frac{10}{3}$, and the semi-major axis is $b = \sqrt{5}$. The length of the latus rectum is given by the formula $\frac{2a^2}{b}$: $$\text{Latus Rectum} = \frac{2(\frac{10}{3})}{\sqrt{5}} = \frac{20}{3\sqrt{5}}$$ Rationalizing the denominator, we arrive at the final result: $$\mathbf{\frac{4\sqrt{5}}{3}}$$

JEE Main 2020 - 2 Sep (Evening)

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: 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 :

Q: Wait, why do we assume the ellipse is vertical?

Great question! We looked at the denominators. Since $\theta \in (0, \frac{\pi}{2})$, we know $\cos^2 \theta a^2$. That confirms it's a vertical ellipse!

Q: Can I just use $e = \sqrt{1 - \frac{a^2}{b^2}}$ for any ellipse?

Be careful! That formula is only valid if $b > a$. If you use it blindly, you might get an imaginary eccentricity or a negative value under the square root. Always compare the denominators first to identify the major axis. In JEE, this is a classic trap to test your conceptual clarity.

Q: Why is the latus rectum formula $\frac{2a^2}{b}$ for a vertical ellipse?

For a horizontal ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ($a>b$), the latus rectum is $\frac{2b^2}{a}$. For a vertical ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ($b>a$), the major axis is along the y-axis, so the length is $\frac{2a^2}{b}$. Think of it as $2 \times \frac{\text{minor}^2}{\text{major}}$. It's all about which axis is longer!

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Before you dive into the algebra, pause and look at the equations. They are not in their standard forms. Your first mission is to rewrite both the hyperbola and the ellipse equations into the canonical forms

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

and

\frac{x ^{2}}{C} + \frac{y ^{2}}{D} = 1

. Only then will the values of

a^{2}

and

b^{2}

reveal themselves.

Hint 2

Hint 2 (The Arsenal): Remember the eccentricity definitions. For a hyperbola,

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

. For an ellipse, you must be careful! Check the denominators. If the denominator under

y^{2}

is larger than the one under

x^{2}

, it is a vertical ellipse, and the formula is

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

. Don't let the orientation trap you.

Hint 3

Hint 3 (The Execution): Once you have

e_{H} = 1 + cos^{2} θ

and

e_{E} = sin θ

, equate them using the given condition

e_{H} = 5 e_{E}

. Square both sides to eliminate the radicals and use the identity

sin^{2} θ = 1 - cos^{2} θ

to solve for

cos^{2} θ

. From there, the latus rectum is just a formula away.

Visualized Solution

Standardizing the Hyperbola

Given Hyperbola: $x^{2} - y^{2} sec^{2} θ = 10$
Divide by $10$ : $\frac{x ^{2}}{10} - \frac{y ^{2} s e c ^{2} θ}{10} = 1$
Standard Form: $\frac{x ^{2}}{10} - \frac{y ^{2}}{10 c o s ^{2} θ} = 1$

Eccentricity of Hyperbola $e_{H}$

Formula: $e_{H} = 1 + \frac{b ^{2}}{a ^{2}}$
Substitute: $e_{H} = 1 + \frac{10 c o s ^{2} θ}{10}$
Result: $e_{H} = 1 + cos^{2} θ$

Standardizing the Ellipse

Given Ellipse: $x^{2} sec^{2} θ + y^{2} = 5$
Divide by $5$ : $\frac{x ^{2} s e c ^{2} θ}{5} + \frac{y ^{2}}{5} = 1$
Standard Form: $\frac{x ^{2}}{5 c o s ^{2} θ} + \frac{y ^{2}}{5} = 1$

Identifying the Major Axis

Given: $θ \in (0, \frac{π}{2})$
Therefore, $cos^{2} θ < 1$
Comparing denominators: $5 cos^{2} θ < 5$
Conclusion: Vertical Ellipse ( $a^{2} < b^{2}$ )

Eccentricity of Ellipse $e_{E}$

Formula for Vertical Ellipse: $e_{E} = 1 - \frac{a ^{2}}{b ^{2}}$
Substitute: $e_{E} = 1 - \frac{5 c o s ^{2} θ}{5}$
Result: $e_{E} = 1 - cos^{2} θ = sin θ$

The Eccentricity Relation

Given Condition: $e_{H} = 5 e_{E}$
Substitute expressions: $1 + cos^{2} θ = 5 sin θ$

Solving for $θ$

Square both sides: $1 + cos^{2} θ = 5 sin^{2} θ$
Use identity: $sin^{2} θ = 1 - cos^{2} θ$
Equation becomes: $1 + cos^{2} θ = 5 (1 - cos^{2} θ)$

Finding $cos^{2} θ$

Expand: $1 + cos^{2} θ = 5 - 5 cos^{2} θ$
Rearrange terms: $6 cos^{2} θ = 4$
Result: $cos^{2} θ = \frac{2}{3}$

Ellipse Parameters

Recall Ellipse denominators: $a^{2} = 5 cos^{2} θ$ and $b^{2} = 5$
Substitute $cos^{2} θ = \frac{2}{3}$
Calculate $a^{2}$ : $a^{2} = 5 \times \frac{2}{3} = \frac{10}{3}$

Latus Rectum Formula

For a vertical ellipse, Latus Rectum $L R = \frac{2 a ^{2}}{b}$
Substitute $a^{2} = \frac{10}{3}$ and $b = 5$
Expression: $L R = \frac{2 ( \frac{10}{3} )}{5}$

Final Calculation

Simplify numerator: $L R = \frac{\frac{20}{3}}{5}$
Rationalize: $L R = \frac{20}{3 5} \times \frac{5}{5}$
Final Answer: $L R = \frac{4 5}{3}$

00:00 / 00:00

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

Solution Diagram

Analyzing the Hyperbola

The given equation for the hyperbola is

x^{2} - y^{2} sec^{2} θ = 10

. Dividing by

10

, we obtain:

\frac{x ^{2}}{10} - \frac{y ^{2} sec ^{2} θ}{10} = 1

Using the identity

\frac{1}{s e c ^{2} θ} = cos^{2} θ

, the equation simplifies to:

\frac{x ^{2}}{10} - \frac{y ^{2}}{10 cos ^{2} θ} = 1

The eccentricity of a hyperbola is defined by

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

. With

a^{2} = 10

and

b^{2} = 10 cos^{2} θ

, we find:

e_{H} = 1 + \frac{10 cos ^{2} θ}{10} = 1 + cos^{2} θ

Analyzing the Ellipse

The equation for the ellipse is

x^{2} sec^{2} θ + y^{2} = 5

. Dividing by

5

, we get:

\frac{x ^{2} sec ^{2} θ}{5} + \frac{y ^{2}}{5} = 1 \Rightarrow \frac{x ^{2}}{5 cos ^{2} θ} + \frac{y ^{2}}{5} = 1

Since

θ \in (0, \frac{π}{2})

, we know

cos^{2} θ < 1

, which implies

5 cos^{2} θ < 5

. Because the denominator under

y^{2}

is larger, the major axis is vertical.

For a vertical ellipse, the eccentricity is

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

, where

a^{2}

is the smaller denominator. Thus:

e_{E} = 1 - \frac{5 cos ^{2} θ}{5} = 1 - cos^{2} θ = sin θ

Solving for the Parameter

We are given the relationship

e_{H} = 5 e_{E}

. Substituting our expressions, we get:

1 + cos^{2} θ = 5 sin θ

Squaring both sides yields

1 + cos^{2} θ = 5 sin^{2} θ

. Using the identity

sin^{2} θ = 1 - cos^{2} θ

1 + cos^{2} θ = 5 (1 - cos^{2} θ)

1 + cos^{2} θ = 5 - 5 cos^{2} θ \Rightarrow 6 cos^{2} θ = 4 \Rightarrow cos^{2} θ = \frac{2}{3}

Final Calculation

For our vertical ellipse, the semi-minor axis squared is

a^{2} = 5 cos^{2} θ = 5 (\frac{2}{3}) = \frac{10}{3}

, and the semi-major axis is

b = 5

The length of the latus rectum is given by the formula

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

Latus Rectum = \frac{2 ( \frac{10}{3} )}{5} = \frac{20}{3 5}

Rationalizing the denominator, we arrive at the final result:

\frac{4 5}{3}

Common Traps & FAQs

Wait, why do we assume the ellipse is vertical?

Great question! We looked at the denominators. Since

θ \in (0, \frac{π}{2})

, we know

cos^{2} θ < 1

. Thus,

5 cos^{2} θ < 5

. The denominator under

x^{2}

is smaller than the denominator under

y^{2}

. In an ellipse equation

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

, the major axis is always under the larger denominator. Here,

b^{2} = 5

and

a^{2} = 5 cos^{2} θ

, so

b^{2} > a^{2}

. That confirms it's a vertical ellipse!

Can I just use

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

for any ellipse?

Be careful! That formula is only valid if

b > a

. If you use it blindly, you might get an imaginary eccentricity or a negative value under the square root. Always compare the denominators first to identify the major axis. In JEE, this is a classic trap to test your conceptual clarity.

Why is the latus rectum formula

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

for a vertical ellipse?

For a horizontal ellipse

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

(

a > b

), the latus rectum is

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

. For a vertical ellipse

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

(

b > a

), the major axis is along the y-axis, so the length is

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

. Think of it as

2 \times \frac{minor ^{2}}{major}

. It's all about which axis is longer!

Animated Solution for Mathematics - Conic Sections: 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

Standardizing the Hyperbola

Eccentricity of Hyperbola $e_{H}$

Standardizing the Ellipse

Identifying the Major Axis

Eccentricity of Ellipse $e_{E}$

The Eccentricity Relation

Solving for $θ$

Finding $cos^{2} θ$

Ellipse Parameters

Latus Rectum Formula

Final Calculation

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

Analyzing the Hyperbola

Analyzing the Ellipse

Solving for the Parameter

Final Calculation

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 :

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:-

In an ellipse, with centre at the origin, if the difference of the lengths of major axis and minor axis is 10 and one of the foci is at $(0, 53)$ , 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

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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

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 $S$ and $S^{'}$ be the foci of the ellipse and $B$ be any one of the extremities of its minor axis. If $S^{'} B S$ is a right angled triangle with right angle at $B$ and area $(S^{'} B S) = 8$ sq. units, then the length of a 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 :

Select Answer:

Visualized Solution

Standardizing the Hyperbola

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

Standardizing the Ellipse

Identifying the Major Axis

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

The Eccentricity Relation

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

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cos2θ

Ellipse Parameters

Latus Rectum Formula

Final Calculation

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

Analyzing the Hyperbola

Analyzing the Ellipse

Solving for the Parameter

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:-

.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

.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 :

Eccentricity of Hyperbola $e_{H}$

Eccentricity of Ellipse $e_{E}$

Solving for $θ$

Finding $cos^{2} θ$

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:-

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