Why did we get $\cos^2\theta$ for $b_h^2$? Is that always the case?

In this specific problem, the trigonometric identity $\csc^2\theta - 1 = \cot^2\theta$ simplified the expression beautifully. When you multiply $\sin^2\theta$ by $\cot^2\theta$, the $\sin^2\theta$ terms cancel out, leaving you with $\cos^2\theta$. It is not a universal rule, but rather a consequence of the specific parameters given in this problem. Always trust your algebraic simplification!

JEE Advanced 2007

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: A hyperbola, having the transverse axis of length $2 sin θ$ , is confocal with the ellipse $3 x^{2} + 4 y^{2} = 12$ . Then its equation is

Select Answer:

Visualized Solution

Analyzing the Ellipse Equation

Given Ellipse: $3 x^{2} + 4 y^{2} = 12$
Divide both sides by $12$ to convert to standard form:
$\frac{x ^{2}}{4} + \frac{y ^{2}}{3} = 1$
Comparing with $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ :
$a^{2} = 4 ⟹ a = 2$
$b^{2} = 3 ⟹ b = 3$

Calculating Ellipse Eccentricity

Formula for eccentricity ( $e$ ) of an ellipse:
$e^{2} = 1 - \frac{b ^{2}}{a ^{2}}$
Substitute $a^{2} = 4$ and $b^{2} = 3$ :
$e^{2} = 1 - \frac{3}{4} = \frac{1}{4}$
Taking the positive square root:
$e = \frac{1}{2}$

Locating the Foci of the Ellipse

The coordinates of the foci of an ellipse are given by:
$(\pm a e, 0)$
Substitute $a = 2$ and $e = \frac{1}{2}$ :
$(\pm 2 \cdot \frac{1}{2}, 0) = (\pm 1, 0)$
So, the foci are $S (1, 0)$ and $S^{'} (- 1, 0)$

Understanding the Confocal Condition

The hyperbola is confocal with the ellipse.
This means they share the exact same foci.
Foci of the hyperbola: $(\pm a_{h} e_{h}, 0) = (\pm 1, 0)$
Therefore, we establish the relation:
$a_{h} e_{h} = 1$

Analyzing the Transverse Axis

Given length of the transverse axis of the hyperbola:
$2 a_{h} = 2 sin θ$
Dividing by $2$ gives the semi-transverse axis:
$a_{h} = sin θ$

Finding Hyperbola Eccentricity $e_{h}$

Using the confocal relation: $a_{h} e_{h} = 1$
Substitute $a_{h} = sin θ$ :
$sin θ \cdot e_{h} = 1$
Solving for $e_{h}$ :
$e_{h} = \frac{1}{s i n θ} = csc θ$

Calculating $b_{h}^{2}$ for the Hyperbola

For a hyperbola, the relation between semi-axes is:
$b_{h}^{2} = a_{h}^{2} (e_{h}^{2} - 1)$
Substitute $a_{h} = sin θ$ and $e_{h} = csc θ$ :
$b_{h}^{2} = sin^{2} θ (csc^{2} θ - 1)$
Using the trigonometric identity $csc^{2} θ - 1 = cot^{2} θ$ :
$b_{h}^{2} = sin^{2} θ \cdot cot^{2} θ = cos^{2} θ$

Constructing the Final Equation

The standard equation of the hyperbola is:
$\frac{x ^{2}}{a _{h}^{2}} - \frac{y ^{2}}{b _{h}^{2}} = 1$
Substitute $a_{h}^{2} = sin^{2} θ$ and $b_{h}^{2} = cos^{2} θ$ :
$\frac{x ^{2}}{s i n ^{2} θ} - \frac{y ^{2}}{c o s ^{2} θ} = 1$
Which simplifies to:
$x^{2} csc^{2} θ - y^{2} sec^{2} θ = 1$

Summary of Key Concepts

Confocal Conics: Share the same foci coordinates $(\pm a e, 0)$ .
Ellipse Relation: $b^{2} = a^{2} (1 - e^{2})$
Hyperbola Relation: $b^{2} = a^{2} (e^{2} - 1)$
Correct Option: (1)

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

Solution Diagram

Analyzing the Setup

The problem asks us to find the equation of a hyperbola that is confocal with the ellipse defined by

3 x^{2} + 4 y^{2} = 12

. Two curves are confocal if they share the same foci.

First, we normalize the ellipse equation by dividing by

12

\frac{x ^{2}}{4} + \frac{y ^{2}}{3} = 1

Comparing this to the standard form

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

, we identify

a^{2} = 4

and

b^{2} = 3

Unveiling the Foci

For the ellipse, the eccentricity

e

is determined by the relation:

e^{2} = 1 - \frac{b ^{2}}{a ^{2}} = 1 - \frac{3}{4} = \frac{1}{4}

This yields

e = \frac{1}{2}

. The foci are located at

(\pm a e, 0)

, which calculates to:

(\pm 2 \cdot \frac{1}{2}, 0) = (\pm 1, 0)

These coordinates,

(\pm 1, 0)

, are the shared foci for our hyperbola.

The Hyperbola's Identity

Let the hyperbola have the standard form

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

. We are given that the length of the transverse axis is

2 sin θ

, so

2 a_{h} = 2 sin θ

, which implies

a_{h} = sin θ

Since the hyperbola shares the foci

(\pm 1, 0)

, its focal distance

a_{h} e_{h}

must equal

1

a_{h} e_{h} = 1 \Rightarrow e_{h} = \frac{1}{a _{h}} = \frac{1}{sin θ} = csc θ

The Final Synthesis

To find

b_{h}^{2}

, we use the hyperbola property

b_{h}^{2} = a_{h}^{2} (e_{h}^{2} - 1)

. Substituting our expressions for

a_{h}

and

e_{h}

b_{h}^{2} = sin^{2} θ (csc^{2} θ - 1)

Using the trigonometric identity

csc^{2} θ - 1 = cot^{2} θ

, we simplify:

b_{h}^{2} = sin^{2} θ \cdot cot^{2} θ = sin^{2} θ \cdot \frac{cos ^{2} θ}{sin ^{2} θ} = cos^{2} θ

Substituting

a_{h}^{2} = sin^{2} θ

and

b_{h}^{2} = cos^{2} θ

into the standard hyperbola equation, we obtain:

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

The final equation of the hyperbola is $x^{2} csc^{2} θ - y^{2} sec^{2} θ = 1$ .

Common Traps & FAQs

Wait, why do we use different formulas for

b^{2}

in ellipses and hyperbolas?

That is a fantastic observation! The difference arises from the geometric nature of the curves. An ellipse is a bounded, closed curve, so its semi-minor axis

b

is smaller than the semi-major axis

a

, leading to

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

. A hyperbola, however, is unbounded, and its conjugate axis

b

relates to the transverse axis

a

through

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

. Always keep these distinct to avoid sign errors!

Can I just use the focus formula

(\pm a e, 0)

for both the ellipse and the hyperbola?

Yes, absolutely! The coordinate formula for the foci of a standard horizontal conic is

(\pm a e, 0)

. The crucial part is remembering that the value of

e

(eccentricity) is different for the ellipse and the hyperbola. You must calculate

e

for each conic separately before finding the coordinates.

Why did we get

cos^{2} θ

for

b_{h}^{2}

? Is that always the case?

In this specific problem, the trigonometric identity

csc^{2} θ - 1 = cot^{2} θ

simplified the expression beautifully. When you multiply

sin^{2} θ

cot^{2} θ

, the

sin^{2} θ

terms cancel out, leaving you with

cos^{2} θ

. It is not a universal rule, but rather a consequence of the specific parameters given in this problem. Always trust your algebraic simplification!

Animated Solution for Mathematics - Conic Sections: A hyperbola, having the transverse axis of length $2 sin θ$ , is confocal with the ellipse $3 x^{2} + 4 y^{2} = 12$ . Then its equation is

Select Answer:

Visualized Solution

Analyzing the Ellipse Equation

Calculating Ellipse Eccentricity

Locating the Foci of the Ellipse

Understanding the Confocal Condition

Analyzing the Transverse Axis

Finding Hyperbola Eccentricity $e_{h}$

Calculating $b_{h}^{2}$ for the Hyperbola

Constructing the Final Equation

Summary of Key Concepts

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

Analyzing the Setup

Unveiling the Foci

The Hyperbola's Identity

The Final Synthesis

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Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A hyperbola, having the transverse axis of length 2sinθ, is confocal with the ellipse 3x2+4y2=12. Then its equation is

Select Answer:

Visualized Solution

Analyzing the Ellipse Equation

Calculating Ellipse Eccentricity

Locating the Foci of the Ellipse

Understanding the Confocal Condition

Analyzing the Transverse Axis

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

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } bh2​ for the Hyperbola

Constructing the Final Equation

Summary of Key Concepts

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

Analyzing the Setup

Unveiling the Foci

The Hyperbola's Identity

The Final Synthesis

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Finding Hyperbola Eccentricity $e_{h}$

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