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Question
If one of the lines given by ax2 + 2hxy + by2 = 0 is perpendicular to px + qy = 0, show that ap2 + 2hpq + bq2 = 0.
Solution
To prove: ap2 + 2hpq + bq2 = 0.
Let the slope of the pair of straight lines ax2 + 2hxy + by2 = 0 be m1 and m2
Then, m1 + m2 = `(-2"h")/"b"` and m1m2 = `"a"/"b"`
Slope of the line px + qy = 0 is `(-"p")/"q"`
But one of the lines of ax2 + 2hxy + by2 = 0 is perpendicular to px + qy = 0
`=> "m"_1 = "q"/"p"`
Now, m1 + m2 = `(-2"h")/"b"` and m1m2 = `"a"/"b"`
`=> "q"/"p" + "m"_2 = (-2"h")/"b"` and `("q"/"p")"m"_2 = "a"/"b"`
`=> "q"/"p" + "m"_2 = (-2"h")/"b"` and `"m"_2 = "ap"/"bq"`
`=> "q"/"p" + "ap"/"bq" = (-2"h")/"b"`
`=> ("bq"^2 + "ap"^2)/"pq" = - 2"h"`
`=> "bq"^2 + "ap"^2 = - 2"h pq"`
`=> "ap"^2 + 2"hpq" + "bq"^2 = 0`
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