If one of the lines given by ax^{2} + 2hxy + by^{2} = 0 is perpendicular to px + qy = 0, show that ap^{2} + 2hpq + bq^{2} = 0.

#### Solution

**To prove:** ap^{2} + 2hpq + bq^{2} = 0.

Let the slope of the pair of straight lines ax^{2} + 2hxy + by^{2} = 0 be m_{1} and m_{2}

Then, m_{1} + m_{2} = `(-2"h")/"b"` and m_{1}m_{2} = `"a"/"b"`

Slope of the line px + qy = 0 is `(-"p")/"q"`

But one of the lines of ax^{2} + 2hxy + by^{2} = 0 is perpendicular to px + qy = 0

`=> "m"_1 = "q"/"p"`

Now, m_{1} + m_{2} = `(-2"h")/"b"` and m_{1}m_{2} = `"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`