Problem 18 - A Complete Resource Book in Mathematics for JEE Main 2017

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 correct answer is (B) \(e = \frac{2}{\sqrt{3}}\).

Step 1: Understand the Problem

Given a hyperbola equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), point \((x_1, y_1)\) and \((x_2, y_2)\) are endpoints of the double ordinate \(PQ\). The triangle \(OPQ\) is equilateral, where \(O\) is the center \((0,0)\) of the hyperbola. We need to find the eccentricity \(e\) of the hyperbola.

Step 2: Identify Properties of Double Ordinates

Since \(P\) and \(Q\) lie on the same ordinate, by symmetry of the hyperbola, \(PQ\) is parallel to the \(y\)-axis, i.e., \(x_1 = x_2 = x_0\). Substitute into the hyperbola equation to get intersection points \((x_0, \pm y_0)\).

Step 3: Formulate Hypotenuse Condition

For \(\triangle OPQ\) to be equilateral, the hypotenuse of \(\triangle OPQ\) (\(PQ\)) must satisfy the condition \(PQ = OP = OQ\). Thus, \(PQ = \sqrt{(0 - 0)^2 + (2y_0)^2} = 2y_0 = \sqrt{x_0^2 + y_0^2}\). Therefore, \(4y_0^2 = x_0^2 + y_0^2\).

Step 4: Solve for Eccentricity Condition

Substituting \(y_0^2 = b^2(1 + \frac{x_0^2}{a^2})\) from the hyperbola equation and rearranging, we solve \((3b^2 = x_0^2)\) leading to \(x_0^2 = 3b^2\). Then, substituting back gives \(a^2 = 3b^2\) which simplifies \(e^2 = \frac{a^2 + b^2}{a^2} = \frac{4b^2}{3b^2} = \frac{4}{3}\).

Step 5: Find Eccentricity

The eccentricity \(e\) is \(\sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}\), which matches option \(B\).

Key Concepts

Eccentricity

Eccentricity is a crucial concept when studying conic sections such as ellipses, parabolas, and hyperbolas. It tells us how much a conic section deviates from being circular. In the context of a hyperbola, eccentricity (denoted as \(e\)) indicates how "stretched" the hyperbola is.
For a hyperbola, eccentricity \(e\) is always greater than one. This is due to its two distinct branches, which extend infinitely. The formula to compute the eccentricity of a hyperbola is given by:

• \[ e = \sqrt{1 + \frac{b^2}{a^2}} \]

where \(a\) and \(b\) are the semi-major and semi-minor axes, respectively. Specifically for a hyperbola defined by \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), \(a^2\) is always subtracted from \(b^2\).
In our problem, through a series of logical deductions, we concluded the eccentricity \(e = \frac{2}{\sqrt{3}}\). This value fits option B, ensuring that the hyperbola indeed has the right eccentricity for the given conditions, especially when forming an equilateral triangle which is crucial to the solution.

Equilateral Triangle

An equilateral triangle is a special type of triangle where all three sides are equal in length and all angles are 60 degrees. This property makes it unique and symmetrical, providing helpful geometrical insights.
In the exercise, the triangle \(OPQ\) is equilateral, which implies:

• \(OP = OQ = PQ\)
• All angles are 60 degrees

To maintain these properties, specific conditions must be satisfied in terms of the lengths of the sides connected by the vertices. For the hyperbola to support such a triangle with its center \(O\), it ensures that the relationship between its axes and eccentricity adheres to the property of equilateral specifics. This directly influences how we calculate eccentricity and satisfy the conditions laid out by the hyperbola equation, leading us to find \(e = \frac{2}{\sqrt{3}}\) as explained in the calculation above.

Double Ordinate

A double ordinate in the context of hyperbolas refers to a line segment perpendicular to the transverse axis, passing through the curve at two points. This is akin to a "width" measurement along the constant \(x\)-coordinate.
The endpoints of this double ordinate are important as they lie on the hyperbola, and their \(y\)-coordinates differ by equal magnitudes, resulting in a reflection symmetry. If you consider a hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the double ordinate can be expressed as:

• Points \((x_0, y_0)\) and \((x_0, -y_0)\)

This generates significant geometric properties, particularly when forming shapes such as equilateral triangles. Determi
ning these points allows one to derive relationships and calculate properties, such as eccentricity. The symmetry note
d in a double ordinate reflects the inherent symmetry of hyperbolas.