Question

If \(PQ\) is a double ordinate of the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) such that \(OPQ\) is an equilateral triangle, \(O\) being the centre of the hyperbola, then the eccentricity \(e\) of the hyperbola satisfies

• A. \(1<e<\frac{2}{\sqrt{3}}\)
• B. \(e = \frac{2}{\sqrt{3}}\)
• C. \(e = \frac{\sqrt{3}}{2}\)
• D. \(e>\frac{2}{\sqrt{3}}\)

Hint:

Use parametric coordinates and equilateral triangle property.

Answer 1

The Correct Option is D

To solve the given problem, let us first understand the properties of the hyperbola and the condition given:

The hyperbola given is in the standard form:

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

Here, \(O\) is the center of the hyperbola. A double ordinate to the hyperbola is a line segment perpendicular to the transverse axis that passes twice through the hyperbola.

It is given that \(OPQ\) forms an equilateral triangle with \(O\) as the center of hyperbola and \(PQ\) as the double ordinate. Let \( (x_1, y_1) \) and \((x_1, -y_1)\) be the coordinates of \(P\) and \(Q\) respectively because they are symmetrical about the transverse axis.

Given that \( \triangle OPQ\) is equilateral:

• This implies that \( OP = PQ = OQ\).

The coordinates of \(O\) are \((0,0)\), thus the distance criteria simplified for both \(OP\) and \(PQ\) results in:

\(OP = \sqrt{x_1^2 + y_1^2}\)\(PQ = |2y_1|\)

Since \( OP = PQ\), equate the above:

\(\sqrt{x_1^2 + y_1^2} = 2|y_1|\)

Squaring both sides gives:

\(x_1^2 + y_1^2 = 4y_1^2 \implies x_1^2 = 3y_1^2\)

Now substitute the values derived above in the hyperbola equation:

\(\frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} = 1\)

Substitute \( x_1^2 = 3y_1^2 \):

\(\frac{3y_1^2}{a^2} - \frac{y_1^2}{b^2} = 1\)

Simplify further:

\(\frac{3}{a^2} - \frac{1}{b^2} = \frac{1}{y_1^2}\)

Utilizing the property of eccentricity \( e \) of a hyperbola:

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

From the condition that \(\frac{3}{a^2} = \frac{1}{y_1^2} + \frac{1}{b^2}\), we understand that for these conditions to satisfy hyperbolic criteria, \( e \) must satisfy:

• Since \(e = \sqrt{1 + \frac{b^2}{a^2}} \Rightarrow e^2 = \frac{a^2 + b^2}{a^2}\), when \(a^2 \to 3y_1^2\), this implies
  • \( 1 + \frac{b^2}{a^2}\) pushes closer to a certain boundary for equilateral formation.

Checking possible solution \(e > \frac{2}{\sqrt{3}}\) matches, exerting \( e = 2/\sqrt{3} \), equilateral solutions sit ideal this boundary.

Therefore, this satisfies:

• The correct solution is \( e > \frac{2}{\sqrt{3}}\)