Question

If the line x – 1 = 0 is a directrix of the hyperbola kx² – y² = 6, then the hyperbola passes through the point

• A. \((-2\sqrt5,6)\)
• B. \((-\sqrt5,3)\)
• C. \((\sqrt5,-2)\)
• D. \((2\sqrt5,3\sqrt6)\)

Answer 1

The Correct Option is C

To solve this problem, we need to find the parameters of the hyperbola given in the question and then determine which of the points given in the options lies on the hyperbola.

The given equation of the hyperbola is:

\(kx^2 - y^2 = 6\)

We are also given that the line \(x - 1 = 0\) is a directrix of the hyperbola. A hyperbola typically takes the form:

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

For a hyperbola, the directrix is at \(x = \pm \frac{a^2}{e}\), where \(e\) is the eccentricity, and \(e > 1\).

Given the directrix \(x = 1\), we identify:

\(\frac{a^2}{e} = 1\) → \(a^2 = e\)

The general expression for the eccentricity of a hyperbola is given by:

\(e = \frac{c}{a}\),

where \(c^2 = a^2 + b^2\).

Thus, if \(e = a^2\), then:

\(a^2 = e = \frac{c}{a} = a\), hence, \(c = a^3\).

Let's substitute back into the original hyperbola equation \(kx^2 - y^2 = 6\), completing square to check for one of the given points:

1. \(x = \sqrt{5}\), \(y = -2\)
   1. \(k(\sqrt{5})^2 - (-2)^2 = 6\)
   2. \(5k - 4 = 6 \Rightarrow 5k = 10\)
   3. \(k = 2\)

This shows that the point \((\sqrt{5}, -2)\) satisfies the hyperbola equation when \(k = 2\).

Hence, the correct option is \((\sqrt{5}, -2)\).