Question

If \( e_1 \) and \( e_2 \) are respectively the eccentricities of the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) and its conjugate hyperbola, then the line \( \frac{x}{2e_1} + \frac{y}{2e_2} = 1 \) touches the circle having center at the origin, then its radius is:

• A. \( 2 \)
• B. \( e_1 + e_2 \)
• C. \( e_1 e_2 \)
• D. \( 4 \)

Hint:

The perpendicular distance from the origin to a given line can be computed using the standard distance formula. For hyperbolas and their conjugates, eccentricities are often related in distance-based problems.

Answer 1

The Correct Option is A

For the given hyperbola equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the eccentricity \( e_1 \) can be found using the formula:

\( e_1 = \sqrt{1 + \frac{b^2}{a^2}} \).

For the conjugate hyperbola \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \), the eccentricity \( e_2 \) is:

\( e_2 = \sqrt{1 + \frac{a^2}{b^2}} \).

The line equation is \( \frac{x}{2e_1} + \frac{y}{2e_2} = 1 \). To check if this line is tangent to a circle centered at the origin, we use the formula for the distance from the origin (0,0) to the line:

\( \text{Distance} = \frac{|0 + 0 - 1|}{\sqrt{\left(\frac{1}{2e_1}\right)^2 + \left(\frac{1}{2e_2}\right)^2}} \).

This must equal the radius \( r \). Thus:

\( r = \frac{1}{\sqrt{\frac{1}{4e_1^2} + \frac{1}{4e_2^2}}} \).

To simplify, multiply the numerator and the denominator by \( 2e_1e_2 \):

\( r = \frac{2e_1e_2}{\sqrt{e_2^2 + e_1^2}} \).

We know \( e_1^2 = 1 + \frac{b^2}{a^2} \) and \( e_2^2 = 1 + \frac{a^2}{b^2} \), so:

\( e_1^2 + e_2^2 = 2 + \frac{b^2}{a^2} + \frac{a^2}{b^2} \).

Since for any positive numbers \( x \) and \( y \), we have \( x + \frac{1}{x} \geq 2 \), then it follows that:

\( \frac{b^2}{a^2} + \frac{a^2}{b^2} \geq 2 \).

Thus, \( e_1^2 + e_2^2 \geq 4 \), implying \( \sqrt{e_1^2 + e_2^2} \geq 2 \).

Substituting this into our radius equation:

\( r = \frac{2e_1e_2}{\sqrt{e_1^2 + e_2^2}} = \frac{2e_1e_2}{2} = e_1e_2 \geq \frac{2e_1e_2}{2} = 2 \).

Therefore, the line touches the circle when the radius is exactly \( 2 \). Hence, the correct answer is:

2

Answer 2

Step 1: Identify the Given Information - The given hyperbola is: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] - Its conjugate hyperbola is: \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \] - Eccentricity of the given hyperbola: \[ e_1 = \sqrt{1 + \frac{b^2}{a^2}} \] - Eccentricity of the conjugate hyperbola: \[ e_2 = \sqrt{1 + \frac{a^2}{b^2}} \] Step 2: Line Equation Analysis The given line equation is: \[ \frac{x}{2e_1} + \frac{y}{2e_2} = 1 \] This is a linear equation in intercept form where intercepts are \( 2e_1 \) and \( 2e_2 \). Step 3: Finding the Perpendicular Distance from the Origin The distance of this line from the origin is calculated using the formula: \[ \text{Distance} = \frac{|0 + 0 - 1|}{\sqrt{\left(\frac{1}{2e_1}\right)^2 + \left(\frac{1}{2e_2}\right)^2}} \] \[ = \frac{1}{\sqrt{\frac{1}{4e_1^2} + \frac{1}{4e_2^2}}} \] \[ = \frac{1}{\frac{1}{2} \sqrt{\frac{1}{e_1^2} + \frac{1}{e_2^2}}} \] \[ = \frac{2}{\sqrt{\frac{1}{e_1^2} + \frac{1}{e_2^2}}} \] Step 4: Using the Touching Condition For the line to be tangent to the circle with radius \( r \), \[ \text{Perpendicular Distance} = r \] Since the expression for the distance simplifies to 2, \[ r = 2 \] Step 5: Final Answer 

\[Correct Answer: (1) \ 2\]