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Question
If the slope of one of the lines given by ax2 + 2hxy + by2 = 0 is square of the slope of the other line, show that a2b + ab2 + 8h3 = 6abh.
Solution
Let m be the slope of one of the lines given by ax2 + 2hxy + by2 = 0.
Then the other line has slope m2
∴ m + m2 = `(-2"h")/"b"` ....(1) and
(m)(m2) = `"a"/"b"`
i.e. m3 = `"a"/"b"` ....(2)
∴ (m + m2)3 = m3 + (m2)3 + 3(m)(m2)(m + m2) .....[∵ (p + q)3 = p3 + q3 + 3pq(p + q)]
∴ `((- "2h")/"b")^3 = "a"/"b" + "a"^2/"b"^2 + 3"a"/"b"((-"2h")/"b")`
∴ `(-8"h"^3)/"b"^3 = "a"/"b" + "a"^2/"b"^2 - "6ah"/"b"^2`
Multiplying by b3, we get,
- 8h3 = ab2 + a2b - 6abh
∴ a2b + ab2 + 8h3 = 6abh
This is the required condition.
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