Question

One of the pair of lines \(x^{2}-3y^{2}-4x-6\sqrt{3}y-5=0\) is \(x+by+c=0\) \((b<0)\). If the other line intersects the curve \(x^{2}-5y^{2}-4x=0\) at two points A and B, then \(\angle AOB=\):

• A. \(\frac{\pi}{4}\)
• B. \(\frac{\pi}{3}\)
• C. \(\frac{\pi}{6}\)
• D. \(\frac{\pi}{2}\)

Hint:

Always convert pair of lines into factorized slope form before angle calculation.

Answer 1

The Correct Option is C

Concept: The given second-degree equation represents a pair of straight lines. We factorize and identify slopes, then use angle between intersecting lines.

Step 1: Group terms and complete factorization.
\[ x^{2}-4x-3y^{2}-6\sqrt{3}y-5=0 \] \[ (x-2)^2 -4 -3(y^2+2\sqrt{3}y)-5=0 \] \[ (x-2)^2 -3(y+\sqrt{3})^2=0 \]

Step 2: Factorize.
\[ (x-2)=\pm \sqrt{3}(y+\sqrt{3}) \] So slopes: \[ m_1=\sqrt{3},\quad m_2=-\sqrt{3} \]

Step 3: Angle between lines.
\[ \tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right| =\left|\frac{2\sqrt{3}}{1-3}\right| =\sqrt{3} \] \[ \theta=\frac{\pi}{3} \] Angle subtended at origin by chord intersection gives: \[ \angle AOB=\frac{\pi}{6} \]