Question

The distance between the tangents of the hyperbola \( 2x^2 - 3y^2 = 6 \) which are perpendicular to the line \( x - 2y + 5 = 0 \) is:

• A. \( 2\sqrt{2} \)
• B. \( 4 \)
• C. \( \sqrt{2} \)
• D. \( 3\sqrt{2} \)

Hint:

To find the distance between tangents to a conic with given slope, use the tangent equation in slope form. For perpendicularity, use the negative reciprocal of the given line's slope.

Answer 1

The Correct Option is A

Step 1: Find slope of line given 
Given line: \( x - 2y + 5 = 0 \Rightarrow y = \frac{1}{2}x + \frac{5}{2} \), so slope \( m = \frac{1}{2} \) Tangents perpendicular to this line must have slope \( m' = -2 \) (negative reciprocal). 
Step 2: General form of tangent to hyperbola 
General form of tangent to hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) is: \[ y = mx \pm \sqrt{a^2m^2 - b^2} \] Given hyperbola: \( \frac{x^2}{3} - \frac{y^2}{2} = 1 \) (by dividing \( 2x^2 - 3y^2 = 6 \) by 6) So here: \( a^2 = 3, b^2 = 2 \) Put \( m = -2 \) into the formula: \[ y = -2x \pm \sqrt{3(-2)^2 - 2} = -2x \pm \sqrt{12 - 2} = -2x \pm \sqrt{10} \] Step 3: Distance between parallel lines 
Lines are of the form: \( y = -2x + \sqrt{10} \) and \( y = -2x - \sqrt{10} \) Distance between them is: \[ \frac{2\sqrt{10}}{\sqrt{1 + (-2)^2}} = \frac{2\sqrt{10}}{\sqrt{5}} = \frac{2\sqrt{10}}{\sqrt{5}} = 2\sqrt{2} \] % Final Answer Hence, the required distance is \( \boxed{2\sqrt{2}} \).