For hyperbola $\frac{x^2}{\cos^2 \alpha} - \frac{y^2}{\sin^2 \alpha} = 1$ which of the following remains constant with change in '$\alpha$'? Options: (A) abscissae of vertices (B) abscissae of foci (C) eccentricity (D) directrix

### Analyzing the Setup We begin with the given equation of the hyperbola: $$\frac{x^2}{\cos^2 \alpha} - \frac{y^2}{\sin^2 \alpha} = 1$$ This equation represents a family of hyperbolas parameterized by $\alpha$. By comparing this to the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we identify the semi-axes as: $$a^2 = \cos^2 \alpha \quad \text{and} \quad b^2 = \sin^2 \alpha$$ Taking the square roots, we obtain $a = \cos \alpha$ and $b = \sin \alpha$. These values define the geometry of the hyperbola for any given $\alpha$. ### The Eccentricity Bridge To understand the behavior of the foci, we must first determine the eccentricity $e$. The formula for the eccentricity of a hyperbola is: $$e = \sqrt{1 + \frac{b^2}{a^2}}$$ Substituting our expressions for $a^2$ and $b^2$, we get: $$e = \sqrt{1 + \frac{\sin^2 \alpha}{\cos^2 \alpha}}$$ Using the trigonometric identity $\frac{\sin^2 \alpha}{\cos^2 \alpha} = \tan^2 \alpha$, the expression simplifies to: $$e = \sqrt{1 + \tan^2 \alpha} = \sqrt{\sec^2 \alpha} = \sec \alpha$$ ### The Master Equation for Foci The foci of a standard hyperbola are located at $(\pm ae, 0)$. We now calculate the abscissa $x = \pm ae$ using our derived values for $a$ and $e$: $$x = \pm (\cos \alpha)(\sec \alpha)$$ Since $\sec \alpha = \frac{1}{\cos \alpha}$, the product simplifies beautifully: $$x = \pm \left( \cos \alpha \cdot \frac{1}{\cos \alpha} \right) = \pm 1$$ ### Final Conclusion The abscissae of the foci are $x = 1$ and $x = -1$. Because these values are independent of the parameter $\alpha$, the foci remain fixed at the points **$(\pm 1, 0)$**. While other properties such as the vertices $(\pm \cos \alpha, 0)$ and the directrices $x = \pm \cos^2 \alpha$ shift as $\alpha$ varies, the foci represent the **constant truth** within this dynamic system.

JEE Advanced 2003

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: For hyperbola $\frac{x ^{2}}{c o s ^{2} α} - \frac{y ^{2}}{s i n ^{2} α} = 1$ which of the following remains constant with change in ' $α$ '?

Q: Wait, why do we assume $a = \cos \alpha$ and not something else?

Great question! We compare our given equation $\frac{x^2}{\cos^2 \alpha} - \frac{y^2}{\sin^2 \alpha} = 1$ with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. By direct visual mapping, the denominator under $x^2$ is $a^2$, which is $\cos^2 \alpha$. Thus, $a = \cos \alpha$. It is a direct identification, not an assumption!

Select Answer:

Visualized Solution

The Parametric Hyperbola

Given equation: $\frac{x ^{2}}{c o s ^{2} α} - \frac{y ^{2}}{s i n ^{2} α} = 1$
The parameter $α$ changes the shape of the hyperbola.
We need to find which geometric property remains constant.

Standard Form Comparison

Standard equation: $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$
We will compare our given equation with this standard form.

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

Comparing denominators:
$a^{2} = cos^{2} α$
$b^{2} = sin^{2} α$

Semi-Axes Lengths

Taking the square root (assuming first quadrant for lengths):
$a = cos α$
$b = sin α$

Formula for Eccentricity

The eccentricity $e$ determines how "open" the hyperbola is.
Formula: $e = 1 + \frac{b ^{2}}{a ^{2}}$

Setting up Eccentricity

Substitute $a^{2} = cos^{2} α$ and $b^{2} = sin^{2} α$ :
$e = 1 + \frac{s i n ^{2} α}{c o s ^{2} α}$

Calculating Eccentricity

We know that $\frac{s i n α}{c o s α} = tan α$ .
So, $e = 1 + tan^{2} α$
Using trigonometric identity: $1 + tan^{2} α = sec^{2} α$
$e = sec α$

Abscissae of Foci

The foci of a standard hyperbola are located at $(\pm a e, 0)$ .
The x-coordinates (abscissae) are $x = \pm a e$ .

Evaluating $a e$

We found $a = cos α$ and $e = sec α$ .
Substitute these into the abscissae formula:
$x = \pm (cos α) (sec α)$

The Constant Foci

Since $sec α = \frac{1}{c o s α}$ , their product is $1$ .
$x = \pm (cos α) (\frac{1}{c o s α}) = \pm 1$
The foci are at $(\pm 1, 0)$ , which is independent of $α$ .

Are Vertices Constant?

Vertices are at $(\pm a, 0)$ .
Substituting $a = cos α$ , we get $(\pm cos α, 0)$ .
This depends on $α$ , so it is not constant.

Are Directrices Constant?

The equations of directrices are $x = \pm \frac{a}{e}$ .
$x = \pm \frac{c o s α}{s e c α} = \pm cos^{2} α$ .
This also depends on $α$ , so it is not constant.

Final Answer

Eccentricity $e = sec α$ (Variable)
Vertices $x = \pm cos α$ (Variable)
Directrices $x = \pm cos^{2} α$ (Variable)
Foci $x = \pm 1$ (Constant)
Therefore, the abscissae of foci remain constant.

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The Sigma Insight: Foci, Directrices, and Eccentricity

Solution Diagram

Analyzing the Setup

We begin with the given equation of the hyperbola:

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

This equation represents a family of hyperbolas parameterized by

α

. By comparing this to the standard form

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

, we identify the semi-axes as:

a^{2} = cos^{2} α and b^{2} = sin^{2} α

Taking the square roots, we obtain

a = cos α

and

b = sin α

. These values define the geometry of the hyperbola for any given

α

The Eccentricity Bridge

To understand the behavior of the foci, we must first determine the eccentricity

e

. The formula for the eccentricity of a hyperbola is:

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

Substituting our expressions for

a^{2}

and

b^{2}

, we get:

e = 1 + \frac{sin ^{2} α}{cos ^{2} α}

Using the trigonometric identity

\frac{s i n ^{2} α}{c o s ^{2} α} = tan^{2} α

, the expression simplifies to:

e = 1 + tan^{2} α = sec^{2} α = sec α

The Master Equation for Foci

The foci of a standard hyperbola are located at

(\pm a e, 0)

. We now calculate the abscissa

x = \pm a e

using our derived values for

a

and

e

x = \pm (cos α) (sec α)

Since

sec α = \frac{1}{c o s α}

, the product simplifies beautifully:

x = \pm (cos α \cdot \frac{1}{cos α}) = \pm 1

Final Conclusion

The abscissae of the foci are

x = 1

and

x = - 1

. Because these values are independent of the parameter

α

, the foci remain fixed at the points $(\pm 1, 0)$ .

While other properties such as the vertices

(\pm cos α, 0)

and the directrices

x = \pm cos^{2} α

shift as

α

varies, the foci represent the constant truth within this dynamic system.

Common Traps & FAQs

Wait, why do we assume

a = cos α

and not something else?

Great question! We compare our given equation

\frac{x ^{2}}{c o s ^{2} α} - \frac{y ^{2}}{s i n ^{2} α} = 1

with the standard form

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

. By direct visual mapping, the denominator under

x^{2}

a^{2}

, which is

cos^{2} α

. Thus,

a = cos α

. It is a direct identification, not an assumption!

Why is the eccentricity not constant? It feels like it should be.

It is a common intuition to think that if the foci are constant, everything else must be, but that is not the case. The eccentricity

e

depends on the ratio of the semi-axes. Since both

a

and

b

change with

α

, their ratio changes, and consequently,

e

changes. In this problem,

e = sec α

, which clearly varies as

α

changes.

How do I know which property to check first?

In JEE problems involving parameters, always start by expressing the standard parameters (

a, b, e

) in terms of the parameter (

α

). Once you have

a = cos α

b = sin α

, and

e = sec α

, you can easily test any property. The foci are usually the most elegant to check because the

a

and

e

terms often simplify beautifully.

Animated Solution for Mathematics - Conic Sections: For hyperbola $\frac{x ^{2}}{c o s ^{2} α} - \frac{y ^{2}}{s i n ^{2} α} = 1$ which of the following remains constant with change in ' $α$ '?

Select Answer:

Visualized Solution

The Parametric Hyperbola

Standard Form Comparison

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

Semi-Axes Lengths

Formula for Eccentricity

Setting up Eccentricity

Calculating Eccentricity

Abscissae of Foci

Evaluating $a e$

The Constant Foci

Are Vertices Constant?

Are Directrices Constant?

Final Answer

The Sigma Insight: Foci, Directrices, and Eccentricity

Analyzing the Setup

The Eccentricity Bridge

The Master Equation for Foci

Final Conclusion

Similar Questions

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List-I

List-II

Let H be the hyperbola, whose foci are $(1 \pm 2, 0)$ and eccentricity is $2$ . Then the length of its latus rectum is

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Let the eccentricity of the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ be a reciprocal to that of the ellipse $x^{2} + 4 y^{2} = 4$ . If the hyperbola passes through a focus of the ellipse, then

Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For hyperbola cos2αx2​−sin2αy2​=1 which of the following remains constant with change in 'α'?

Select Answer:

Visualized Solution

The Parametric Hyperbola

Standard Form Comparison

Identifying .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

Semi-Axes Lengths

Formula for Eccentricity

Setting up Eccentricity

Calculating Eccentricity

Abscissae of Foci

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

The Constant Foci

Are Vertices Constant?

Are Directrices Constant?

Final Answer

The Sigma Insight: Foci, Directrices, and Eccentricity

Analyzing the Setup

The Eccentricity Bridge

The Master Equation for Foci

Final Conclusion

Similar Questions

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List-I

List-II

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Animated Solution for Mathematics - Conic Sections: For hyperbola $\frac{x ^{2}}{c o s ^{2} α} - \frac{y ^{2}}{s i n ^{2} α} = 1$ which of the following remains constant with change in ' $α$ '?

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

Evaluating $a e$

Let $0 < θ < \frac{π}{2}$ . If the eccentricity of the hyperbola $\frac{x ^{2}}{c o s ^{2} θ} - \frac{y ^{2}}{s i n ^{2} θ} = 1$ is greater than 2, then the length of its latus rectum lies in the interval :

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