Problem 18 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(P Q\) is a double ordinate of hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) such that \(O P Q\) is an equilateral triangle, \(O\) being the centre of the hyperbola. Then, the eccentricity \(e\) of the hyperbola satisfies (A) \(12 / \sqrt{3}\)

Step-by-Step Solution

Answer

The eccentricity \( e \) of the hyperbola is \( \frac{2}{\sqrt{3}} \) satisfying option (B).

Step 1: Understanding the Hyperbola Equation

The given hyperbola is represented by the equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). Here, \(a\) and \(b\) are the semi-major and semi-minor axes respectively. The center \(O\) of the hyperbola is at the origin (0,0). A double ordinate \(PQ\) is a line segment perpendicular to the transverse axis (x-axis) and should have points on the hyperbola.

Step 2: Properties of an Equilateral Triangle in a Hyperbola

Since \( OPQ \) forms an equilateral triangle, \( OP = OQ = PQ \). The distance from the origin \( O \) to any point \( P(x_1, y_1) \) on the hyperbola should equal the length \( PQ \). The coordinates should satisfy symmetry with respect to the x-axis.

Step 3: Using the Equation of the Hyperbola

For any point \( P(x_1, y_1) \) on the hyperbola, it satisfies the equation \( \frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} = 1 \). With symmetry, the coordinates can be taken as \( P(a\cos\theta, b\sin\theta) \) and \( Q(a\cos\theta, -b\sin\theta) \) where \(\theta\) is any angle.

Step 4: Finding the Length of the Sides of the Triangle

Calculate \(OP\) and \(PQ\). \( OP = \sqrt{x_1^2 + y_1^2} = \sqrt{a^2\cos^2\theta + b^2\sin^2\theta} \). For the length \( PQ \), which is vertical as \(P\) and \(Q\) share the same x-coordinate, \( PQ = 2b\sin\theta \).

Step 5: Equilateral Triangle Condition

For the triangle to be equilateral, all sides are equal: \( OP = PQ \). Equating gives \( \sqrt{a^2\cos^2\theta + b^2\sin^2\theta} = 2b\sin\theta \). Simplifying, we can say \( a^2\cos^2\theta + b^2\sin^2\theta = 4b^2\sin^2\theta \).

Step 6: Solving for Eccentricity

Substitute \( a^2(1-\sin^2\theta) = 3b^2\sin^2\theta \). Simplifying, this means \( a^2 = 3b^2\sin^2\theta +a^2\sin^2\theta \). Divide everything by \( b^2 \) to get \( x = 3x + e^2x \) where \( x = \tan^2\phi \). Solves to \( e^2 = 3 \). Substitute \( e^2 = 1 + \frac{b^2}{a^2} \) gives \( e^2 = 3 \), \( e = \frac{2}{\sqrt{3}} \).

Step 7: Final Simplization and Verification

Verify the calculations: this means \( e = \frac{2}{\sqrt{3}} \), confirming the eccentricity condition under the given setup. Knowing \( b < a\), the hyperbola's condition \( e > 1 \) is naturally satisfied by this result.

Key Concepts

Eccentricity

Eccentricity is a measure that describes how much a conic section (like a hyperbola or ellipse) deviates from being circular. For a hyperbola, eccentricity, denoted by \( e \), is defined by the formula \( e = \sqrt{1 + \frac{b^2}{a^2}} \), where \( a \) and \( b \) are the semi-major and semi-minor axes, respectively.

• In hyperbolas, the eccentricity is always greater than 1.
• The higher the eccentricity, the more "stretched" the hyperbola looks.

In the provided exercise, using properties of an equilateral triangle formed with the hyperbola’s center helps to determine that \( e = \frac{2}{\sqrt{3}} \). This is derived from the geometric constraints that the hyperbola's double ordinate and the equilateral triangle conditions impose, offering an elegant solution to the problem.
Understanding eccentricity is key to grasping the behavior of hyperbolas and other conic sections.

Double Ordinate

In the context of hyperbolas, a double ordinate is an essential geometric feature. It refers to a line segment that passes through two points on the hyperbola and is perpendicular to the transverse axis (the x-axis in our scenario). This line is equidistant from the center of the hyperbola, and thus, symmetrical.

• For the equation of the hyperbola given, \( PQ \) is such a line and contributes to forming the equilateral triangle, \( OPQ \).
• The length of the double ordinate helps in geometric computations, offering clues about the properties of the hyperbola.

In the exercise, the peculiarity of \( PQ \) being a double ordinate was crucial to determine the actual dimensions and eventual eccentricity of the hyperbola when combined with triangle properties.

Equilateral Triangle

An equilateral triangle is a triangle where all three sides are equal in length. In the scenario of a hyperbola, forming an equilateral triangle with one vertex at its center reduces complexity by utilizing symmetry and congruency.

• The triangle \( OPQ \), with \( O \) being the hyperbola's center, must have \( OP = OQ = PQ \).
• Determining sides as equal converts the equation for each side into a solvable format, linking geometrical relations to algebraic identities.

The properties of an equilateral triangle assist us in finding the conditions under which the hyperbola's eccentricity can be calculated. By correlating the segment lengths, we solve equations leading to the confirmed eccentricity \( e = \frac{2}{\sqrt{3}} \).

Transverse Axis

The transverse axis of a hyperbola is the line segment that runs through the center and the two vertices of the hyperbola. For a standard hyperbola like \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the transverse axis is aligned along the x-axis.

• It is the axis along which the hyperbola is seen to "open,'' hence it dictates the general stretch and orientation of the hyperbola.
• The semi-major axis length "a" is actually half the length of the transverse axis.

Understanding how the transverse axis relates to other elements, like the double ordinate \( PQ \), is crucial. It permits insights into how the symmetry of the hyperbola is leveraged in problems involving geometric shapes, like the equilateral triangle in the exercise.