Question

The equation \(x^2 - Ky^2 - 4x + 6y - 5 = 0\) represents a pair of straight lines. Find the point of intersection.

Hint:

Partial differentiation is the fastest way to find the intersection of any second-degree curve (Pair of lines, Circle, etc.) with itself. It avoids solving quadratic equations completely!

Answer 1

Step 1: Understanding the Concept:
The point of intersection of a pair of straight lines \(S(x,y) = 0\) is the point where the partial derivatives \(\frac{\partial S}{\partial x} = 0\) and \(\frac{\partial S}{\partial y} = 0\) vanish.

Step 2: Key Formula or Approach:
1. Condition for pair of lines: \(\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0\).
2. Point of intersection: Solve \(\frac{\partial S}{\partial x} = 0\) and \(\frac{\partial S}{\partial y} = 0\).

Step 3: Detailed Explanation:
Let \(S(x,y) = x^2 - Ky^2 - 4x + 6y - 5 = 0\).
First, find \(K\) using the condition \(\Delta = 0\):
\(a=1, b=-K, c=-5, g=-2, f=3, h=0\).
\[ 1(-K)(-5) + 0 - 1(3)^2 - (-K)(-2)^2 - 0 = 0 \] \[ 5K - 9 + 4K = 0 \implies 9K = 9 \implies K = 1 \] Now find the intersection point using partial derivatives:
\(\frac{\partial S}{\partial x} = 2x - 4 = 0 \implies x = 2\).
\(\frac{\partial S}{\partial y} = -2Ky + 6 = 0\).
Substituting \(K=1\):
\(-2(1)y + 6 = 0 \implies 2y = 6 \implies y = 3\).
The point is \((2,3)\).

Step 4: Final Answer:
The point of intersection is \((2, 3)\).