# If one of the lines given by ax2 + 2hxy + by2 = 0 is perpendicular to px + qy = 0, show that ap2 + 2hpq + bq2 = 0. - Mathematics and Statistics

Sum

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

Concept: Homogeneous Equation of Degree Two
Is there an error in this question or solution?