Problem 221 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

For the Hyperbola \(\frac{x^{2}}{\cos ^{2} \alpha}-\frac{y^{2}}{\sin ^{2} \alpha}=1\), which of the following remains constant when \(\alpha\) varies \(=\) ? (a) abscissae of vertices (b) abscissae of foci (c) eccentricity (d) directrix.

Step-by-Step Solution

Answer

(b) Abscissae of foci remain constant.

Step 1: Understanding the Hyperbola

The equation given is \( \frac{x^{2}}{\cos^{2} \alpha} - \frac{y^{2}}{\sin^{2} \alpha} = 1 \). This is a standard form of a hyperbola equation \( \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \), where \( a = \cos \alpha \) and \( b = \sin \alpha \).

Step 2: Determine Vertices

For a hyperbola of the form \( \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \), the vertices occur at \( (\pm a, 0) \). Since \( a = \cos \alpha \), the abscissae of the vertices are \( \pm \cos \alpha \). These values change as \( \alpha \) changes.

Step 3: Determine Foci

The foci for the hyperbola are given by \( (\pm c, 0) \), where \( c = \sqrt{a^{2} + b^{2}} = \sqrt{\cos^{2} \alpha + \sin^{2} \alpha} \). Since \( \cos^{2} \alpha + \sin^{2} \alpha = 1 \), \( c = 1 \), thus the abscissae of foci are constant \( \pm 1 \).

Step 4: Check Eccentricity

The eccentricity \( e \) of the hyperbola is defined as \( e = \frac{c}{a} \). With \( c = 1 \) and \( a = \cos \alpha \), so \( e = \frac{1}{\cos \alpha} \). Hence, the eccentricity changes as \( \alpha \) changes.

Step 5: Check Directrix

The directrix \( x = \pm \frac{a}{e} \) can also be expressed with \( a = \cos \alpha \) and \( e = \frac{1}{\cos \alpha} \). This simplifies to \( x = \pm \cos^{2} \alpha \), which varies with \( \alpha \).

Key Concepts

Vertices of Hyperbola

In a hyperbola, the vertices are special points located on the major axis. They represent the closest points of the hyperbola to its center. For hyperbolas of the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the vertices are at the coordinates \((\pm a, 0)\). Here, \(a\) is the distance from the center to each vertex along the major axis. In the exercise provided, \( a = \cos \alpha \), leading to vertices at \( (\pm \cos \alpha, 0) \).
Therefore, as \( \alpha \) changes, \( \cos \alpha \) varies, causing the positions of the vertices to change. This explains why the abscissae (the \(x\)-coordinates) of the vertices are not constant when \( \alpha \) changes.

Foci of Hyperbola

The foci of a hyperbola are two important points that define its properties. These points lie on the major axis of the hyperbola and are outside the vertices. The hyperbola's defining characteristic is that the difference in distances from any point on the curve to the two foci is constant.
For a hyperbola in the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the foci are at \((\pm c, 0)\), where \(c\) is given by \(c = \sqrt{a^2 + b^2}\). Using the identity \( \cos^2 \alpha + \sin^2 \alpha = 1 \), it's determined that here \(c = 1\).
This makes the abscissae of the foci constant at \(\pm 1\), regardless of the value of \( \alpha \), since both \( a\) and \( b\) resulted in a \(c\) value unaffected by \( \alpha \).

Eccentricity of Hyperbola

Eccentricity, often denoted by \(e\), is a parameter that describes how much a conic section deviates from being circular. For a hyperbola, eccentricity helps to determine how 'stretched' it is. The formula to calculate the eccentricity of a hyperbola is \( e = \frac{c}{a} \).
In the case of the given hyperbola, \(c = 1\) and \( a = \cos \alpha \), leading to \( e = \frac{1}{\cos \alpha} \). Since \( \cos \alpha \) varies with \( \alpha \), \( e \) also varies. Therefore, the eccentricity changes as \( \alpha \) is altered, indicating that it is not a constant value. This variability distinguishes hyperbolas based on their opening angle and proportion.

Directrix of Hyperbola

A directrix is an important element of a hyperbola's geometry. It's a line used in the definition of the conic section alongside the focus. The directrix assists in determining a hyperbola's eccentricity and other parameters. For a hyperbola, it is typically represented as \(x = \pm \frac{a}{e}\).
Using the given expression, substituting \(a = \cos \alpha\) and \(e = \frac{1}{\cos \alpha}\) leads to \(x = \pm \cos^2 \alpha\). Since \( \cos \alpha\) changes with \( \alpha\), the directrix \(x = \pm \cos^2 \alpha\) inherently varies as \( \alpha\) varies.
In essence, for the hyperbola given, even the directrix is not constant as it depends on the angle \(\alpha\), reflecting how sensitive hyperbolic shapes are to changes in this parameter.