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Question
If the lines given by ax2 + 2hxy + by2 = 0 form an equilateral triangle with the line lx + my = 1, show that (3a + b)(a + 3b) = 4h2.
Solution
Since the lines ax2 + 2hxy + by2 = 0 form an equilateral triangle with the line
lx + my = 1, the angle between the lines
ax2 + 2hxy + by2 = 0 is 60°.
∴ tan 60° = `|(2sqrt("h"^2 - "ab"))/("a + b")|`
∴ `sqrt3 = |(2sqrt("h"^2 - "ab"))/("a + b")|`
∴ 3(a + b)2 = 4(h2 - ab)
∴ 3(a2 + 2ab + b2) = 4h2 - 4ab
∴ 3a2 + 6ab + 3b2 + 4ab = 4h2
∴ 3a2 + 10ab + 3b2 = 4h2
∴ 3a2 + 9ab + ab + 3b2 = 4h2
∴ 3a(a + 3b) + b(a + 3b) = 4h2
∴ (3a + b)(a + 3b) = 4h2
This is the required condition.
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