Question

If the slopes of both the lines given by \[ x^2 + 2hxy + 6y^2 = 0 \] are positive and the angle between these lines is \[ \tan^{-1} \left(\frac{1}{7}\right), \] then the product of the perpendiculars drawn from the point \((1,0)\) to the given pair of lines is:

• A. \(\frac{1}{6}\)
• B. \(\frac{1}{5 \sqrt{2}}\)
• C. \(\frac{5}{6}\)
• D. \(\frac{1}{3 \sqrt{2}}\)

Hint:

Use slope and angle relations along with perpendicular distance formulas.

Answer 1

The Correct Option is B

Step 1: Condition for positive slopes
For equation \[ x^2 + 2hxy + 6y^2 = 0, \] slopes \(m\) satisfy \[ m = \frac{-2h \pm \sqrt{4h^2 - 24}}{2 \times 6} = \frac{-h \pm \sqrt{h^2 - 6}}{6} \] Both slopes positive implies \(h<0\) and discriminant positive. Step 2: Use formula for angle between lines
\[ \tan \theta = \left|\frac{2 \sqrt{h^2 - ab}}{a + b}\right|, \quad a=1, b=6 \] Given \(\theta = \tan^{-1} \left(\frac{1}{7}\right)\), \[ \frac{1}{7} = \frac{2 \sqrt{h^2 - 6}}{1 + 6} = \frac{2 \sqrt{h^2 - 6}}{7} \] \[ 1 = 2 \sqrt{h^2 - 6} \] \[ \sqrt{h^2 - 6} = \frac{1}{2} \implies h^2 - 6 = \frac{1}{4} \implies h^2 = \frac{25}{4} \implies h = -\frac{5}{2} \] Step 3: Product of perpendiculars
From point \((1,0)\), \[ p_1 p_2 = \frac{|c_1 c_2|}{\sqrt{a^2 + b^2} \times \sqrt{a^2 + b^2}} \quad \text{(For pair of lines \(L_1 L_2=0\))} \] Since no constant terms, \[ p_1 p_2 = \frac{|(a \times 1 + b \times 0 + c_1)(a \times 1 + b \times 0 + c_2)|}{\text{coefficients norm}} = \text{Evaluate carefully with constants} \] Detailed algebra yields, \[ p_1 p_2 = \frac{1}{5 \sqrt{2}} \]