Q: A straight line through P (–2 , –3) cuts the pair of straight lines x^{2} + 3y^{2} + 4xy – 8x – 6y – 9 = 0 in Q and R. Find the equation of the line if PQ . PR = 20.

Sol: Let line be $\displaystyle \frac{x+2}{cos\theta} = \frac{y+3}{sin\theta} = r $

x = r cosθ – 2, y = r sinθ – 3 …(i)

Now, x^{2} + 3y^{2} + 4xy – 8x – 6y – 9 = 0 …(ii)

Taking intersection of (i) with (ii) and considering terms of r_{2} and constant (as we need PQ.PR = r_{1}.r_{2} = product of the roots)

r^{2}(cos^{2} θ + 3 sin^{2} θ + 4 sin θ cos θ) + (some terms)r + 80 = 0

$\displaystyle r_1 . r_2 = PQ.PR = \frac{80}{cos^2 \theta + 4 sin\theta cos\theta + 3 sin^2 \theta} $

⇒ cos^{2} θ + 4 sin θ cos θ + 3 sin^{2} θ = 4 (As PQ.PR = 20)

⇒ sin^{2} θ – 4 sin θ cos θ + 3 cos^{2} θ = 0

⇒ (sin θ – cos θ)(sin θ – 3 cos θ) = 0

⇒ tan θ = 1, tan θ = 3

hence equation of the line is y + 3 = 1(x + 2)

⇒ x – y = 1

and y + 3 = 3(x + 2)

⇒ 3x – y + 3 = 0.