The distance between the tangents of the hyperbola \( 2x^2 - 3y^2 = 6 \) which are perpendicular to the line \( x - 2y + 5 = 0 \) is:
Show Hint
- \( 2\sqrt{2} \)
- \( 4 \)
- \( \sqrt{2} \)
- \( 3\sqrt{2} \)
The Correct Option is A
Solution and Explanation
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 \( \boxed{2\sqrt{2}} \).
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