JEE Main 2024 (01 Feb Shift 1)

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Animated Solution for Mathematics - Conic Sections: For $0 < θ < π /2$ , if the eccentricity of the hyperbola $x^{2} - y^{2} cosec^{2} θ = 5$ is $7$ times eccentricity of the ellipse $x^{2} cosec^{2} θ + y^{2} = 5$ , then the value of $θ$ is :

Q: Wait, why did the formula for the ellipse eccentricity change? I thought it was always $\sqrt{1 - b^2/a^2}$.

That is a common trap! The standard formula $e = \sqrt{1 - b^2/a^2}$ assumes $a > b$ (a horizontal ellipse). When the ellipse is vertical (major axis along the y-axis), the formula becomes $\sqrt{1 - \frac{\text{minor axis}^2}{\text{major axis}^2}}$. Always check which denominator is larger before applying the formula.

Q: Why can't we just use energy conservation or some other physics principle here?

This is a pure coordinate geometry problem, not a physics problem! While the language of 'conic sections' is used in physics (like orbits), here we are strictly dealing with the algebraic definitions of hyperbolas and ellipses. Stick to the geometric properties of conics.

Q: Why do we reject the negative root when we take the square root of $\sin^2 \theta = 3/4$?

Excellent observation! We have $\sin \theta = \pm \frac{\sqrt{3}}{2}$. However, the problem explicitly states that $0 < \theta < \frac{\pi}{2}$. In the first quadrant, the sine function is strictly positive. Therefore, we must reject the negative value.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. To solve this, you must first recognize the standard forms of the conic sections. The hyperbola

x^{2} - y^{2} csc^{2} θ = 5

and the ellipse

x^{2} csc^{2} θ + y^{2} = 5

are not in their standard forms. You need to divide by the constant on the right side and express the coefficients in terms of

sin^{2} θ

cos^{2} θ

. Remember, the eccentricity formula for a hyperbola is

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

, but for an ellipse, it depends on which axis is the major axis.

Hint 2

Hint 2: The Arsenal. You must correctly identify the major axis for the ellipse. Since

0 < θ < \frac{π}{2}

, we know that

0 < sin^{2} θ < 1

. This means the denominator under

y^{2}

is larger than the denominator under

x^{2}

. Therefore, the ellipse is vertical. Use the correct eccentricity formula for a vertical ellipse:

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

, where

a^{2}

is the smaller denominator and

b^{2}

is the larger one.

Hint 3

Hint 3: The Execution. Once you have both eccentricities,

e_{h} = 1 + sin^{2} θ

and

e_{e} = cos θ

, set up the equation

e_{h} = 7 e_{e}

. Square both sides to eliminate the radicals:

1 + sin^{2} θ = 7 cos^{2} θ

. Now, convert the entire equation into a single trigonometric function using the identity

cos^{2} θ = 1 - sin^{2} θ

and solve for

θ

Visualized Solution

Problem Setup

Given Hyperbola: $x^{2} - y^{2} csc^{2} θ = 5$
Given Ellipse: $x^{2} csc^{2} θ + y^{2} = 5$
Condition: $e_{h} = 7 e_{e}$
Constraint: $0 < θ < \frac{π}{2}$

Standardizing the Hyperbola

Divide by $5$ : $\frac{x ^{2}}{5} - \frac{y ^{2} c s c ^{2} θ}{5} = 1$
Use $csc^{2} θ = \frac{1}{s i n ^{2} θ}$ :
$\frac{x ^{2}}{5} - \frac{y ^{2}}{5 s i n ^{2} θ} = 1$
Compare with $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$

Hyperbola Eccentricity $e_{h}$

Formula: $e_{h} = 1 + \frac{b ^{2}}{a ^{2}}$
Substitute $a^{2} = 5$ and $b^{2} = 5 sin^{2} θ$ :
$e_{h} = 1 + \frac{5 s i n ^{2} θ}{5}$

Simplifying $e_{h}$

Cancel out the $5$ :
$e_{h} = 1 + sin^{2} θ$

Standardizing the Ellipse

Divide by $5$ : $\frac{x ^{2} c s c ^{2} θ}{5} + \frac{y ^{2}}{5} = 1$
Rewrite using $sin^{2} θ$ :
$\frac{x ^{2}}{5 s i n ^{2} θ} + \frac{y ^{2}}{5} = 1$

Identifying the Major Axis

Compare denominators: $5 sin^{2} θ$ and $5$
Since $0 < θ < \frac{π}{2}$ , $0 < sin^{2} θ < 1$
Therefore, $5 sin^{2} θ < 5$
The major axis is along the y-axis.

Ellipse Eccentricity $e_{e}$

For a vertical ellipse: $e_{e} = 1 - \frac{a ^{2}}{b ^{2}}$
Substitute $a^{2} = 5 sin^{2} θ$ and $b^{2} = 5$ :
$e_{e} = 1 - \frac{5 s i n ^{2} θ}{5}$

Simplifying $e_{e}$

Cancel out the $5$ :
$e_{e} = 1 - sin^{2} θ$
Use identity $1 - sin^{2} θ = cos^{2} θ$ :
$e_{e} = cos^{2} θ = cos θ$

Applying the Given Condition

Given: $e_{h} = 7 e_{e}$
Substitute $e_{h} = 1 + sin^{2} θ$ and $e_{e} = cos θ$ :
$1 + sin^{2} θ = 7 cos θ$

Squaring Both Sides

Square both sides to remove the square root:
$1 + sin^{2} θ = 7 cos^{2} θ$

Converting to a Single Function

Use $cos^{2} θ = 1 - sin^{2} θ$ :
$1 + sin^{2} θ = 7 (1 - sin^{2} θ)$

Solving for $sin^{2} θ$

Expand the right side:
$1 + sin^{2} θ = 7 - 7 sin^{2} θ$
Rearrange terms:
$8 sin^{2} θ = 6$
$sin^{2} θ = \frac{6}{8} = \frac{3}{4}$

Finding the Value of $θ$

Take the square root: $sin θ = \pm \frac{3}{2}$
Since $0 < θ < \frac{π}{2}$ , $sin θ$ must be positive:
$sin θ = \frac{3}{2}$
Therefore, $θ = \frac{π}{3}$

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

Solution Diagram

Analyzing the Hyperbola's Secret

Let us begin with the hyperbola:

x^{2} - y^{2} csc^{2} θ = 5

. To understand its nature, we must bring it into the light of its standard form,

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

Dividing the entire equation by

5

, we get:

\frac{x ^{2}}{5} - \frac{y ^{2} csc ^{2} θ}{5} = 1

Using the identity

csc^{2} θ = \frac{1}{s i n ^{2} θ}

, this transforms into:

\frac{x ^{2}}{5} - \frac{y ^{2}}{5 sin ^{2} θ} = 1

Here,

a^{2} = 5

and

b^{2} = 5 sin^{2} θ

. The eccentricity of a hyperbola is defined as

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

Substituting our values, we find:

e_{h} = 1 + \frac{5 sin ^{2} θ}{5} = 1 + sin^{2} θ

This is our first anchor point.

The Ellipse's Hidden Orientation

Now, turn your gaze to the ellipse:

x^{2} csc^{2} θ + y^{2} = 5

. Again, we divide by

5

to standardize:

\frac{x ^{2} csc ^{2} θ}{5} + \frac{y ^{2}}{5} = 1 \Rightarrow \frac{x ^{2}}{5 sin ^{2} θ} + \frac{y ^{2}}{5} = 1

Here lies the trap! We are given

0 < θ < \frac{π}{2}

, which implies

0 < sin^{2} θ < 1

. Consequently,

5 sin^{2} θ < 5

Since the larger denominator (

5

) is under the

y^{2}

term, 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 (

5 sin^{2} θ

) and

b^{2}

is the larger one (

5

Substituting these, we get:

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

The Grand Synthesis

We have our two eccentricities:

e_{h} = 1 + sin^{2} θ

and

e_{e} = cos θ

. The problem dictates that

e_{h} = 7 e_{e}

Substituting our expressions, we arrive at:

1 + sin^{2} θ = 7 cos θ

To break this radical, we square both sides:

1 + sin^{2} θ = 7 cos^{2} θ

Now, we convert

cos^{2} θ

into

1 - sin^{2} θ

1 + sin^{2} θ = 7 (1 - sin^{2} θ)

Expanding this, we have

1 + sin^{2} θ = 7 - 7 sin^{2} θ

. Rearranging the terms, we find:

8 sin^{2} θ = 6 \Rightarrow sin^{2} θ = \frac{3}{4}

The Final Victory

Taking the square root, we get

sin θ = \pm \frac{3}{2}

. Given our constraint

0 < θ < \frac{π}{2}

, we know

sin θ

must be positive.

Thus,

sin θ = \frac{3}{2}

. The angle that satisfies this in the first quadrant is:

θ = \frac{π}{3}

We have navigated the coordinate plane, identified the orientation of the conics, and solved the trigonometric puzzle. The final result is $θ = \frac{π}{3}$ .

Common Traps & FAQs

Wait, why did the formula for the ellipse eccentricity change? I thought it was always

1 - b^{2} / a^{2}

That is a common trap! The standard formula

e = 1 - b^{2} / a^{2}

assumes

a > b

(a horizontal ellipse). When the ellipse is vertical (major axis along the y-axis), the formula becomes

1 - \frac{minor axis ^{2}}{major axis ^{2}}

. Always check which denominator is larger before applying the formula.

Why can't we just use energy conservation or some other physics principle here?

This is a pure coordinate geometry problem, not a physics problem! While the language of 'conic sections' is used in physics (like orbits), here we are strictly dealing with the algebraic definitions of hyperbolas and ellipses. Stick to the geometric properties of conics.

Why do we reject the negative root when we take the square root of

sin^{2} θ = 3/4

Excellent observation! We have

sin θ = \pm \frac{3}{2}

. However, the problem explicitly states that

0 < θ < \frac{π}{2}

. In the first quadrant, the sine function is strictly positive. Therefore, we must reject the negative value.

Animated Solution for Mathematics - Conic Sections: For $0 < θ < π /2$ , if the eccentricity of the hyperbola $x^{2} - y^{2} cosec^{2} θ = 5$ is $7$ times eccentricity of the ellipse $x^{2} cosec^{2} θ + y^{2} = 5$ , then the value of $θ$ is :

Select Answer:

Visualized Solution

Problem Setup

Standardizing the Hyperbola

Hyperbola Eccentricity $e_{h}$

Simplifying $e_{h}$

Standardizing the Ellipse

Identifying the Major Axis

Ellipse Eccentricity $e_{e}$

Simplifying $e_{e}$

Applying the Given Condition

Squaring Both Sides

Converting to a Single Function

Solving for $sin^{2} θ$

Finding the Value of $θ$

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

Analyzing the Hyperbola's Secret

The Ellipse's Hidden Orientation

The Grand Synthesis

The Final Victory

Similar Questions

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 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 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 a hyperbola passes through the focus of the ellipse $\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1$ . The transverse and conjugate axes of this hyperbola coincide with the major and minor axes of the given ellipse, also the product of eccentricities of given ellipse and hyperbola is 1, then

Let $(h, k)$ lie on the circle $C : x^{2} + y^{2} = 4$ and the point $(2 h + 1, 3 k + 2)$ lie on an ellipse with eccentricity $e$ . Then the value of $\frac{5}{e ^{2}}$ is equal to .........

Select Answer:

Visualized Solution

Problem Setup

Standardizing the Hyperbola

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

Simplifying .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

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

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

Applying the Given Condition

Squaring Both Sides

Converting to a Single Function

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin2θ

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

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

Analyzing the Hyperbola's Secret

The Ellipse's Hidden Orientation

The Grand Synthesis

The Final Victory

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 (h,k) lie on the circle C:x2+y2=4 and the point (2h+1,3k+2) lie on an ellipse with eccentricity e. Then the value of e25​ is equal to .........

Hyperbola Eccentricity $e_{h}$

Simplifying $e_{h}$

Ellipse Eccentricity $e_{e}$

Simplifying $e_{e}$

Solving for $sin^{2} θ$

Finding the Value of $θ$

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 $(h, k)$ lie on the circle $C : x^{2} + y^{2} = 4$ and the point $(2 h + 1, 3 k + 2)$ lie on an ellipse with eccentricity $e$ . Then the value of $\frac{5}{e ^{2}}$ is equal to .........