Asked by clive
x^2 + xy - Ky^2=0 , the angle made by the line pairs of this equation is 45 degree. I got the value of k as 0 and -2, do we need to reject k=0 and conclude our answer as k=-2 ?
Answers
Answered by
Reiny
x^2 + xy - Ky^2=0
To have this equation represented by two straight lines, we must have it in a factored form such as
(x + ay)(x - by) = 0
so that a+b = 1 and ab=-k
e.g. if we had
x^2 + xy - 6y^2 = 0
(x+3y)(x-2y) = 0
our two lines would be
x+3y = 0 or x-2y = 0
slope of first = -1/3
slope of 2nd = 1/2
for acute angle formed:
using the formula
tanθ = (m2 - m1)/(1 + m2m1)
tanθ = (1/2 + 1/3)/(1 + (1/2)(-1/3)
= (5/6) / (5/6)
= 1
θ = 45°
What a lucky guess!
check this graph:
https://www.wolframalpha.com/input/?i=plot+x%5E2+%2B+xy+-+6y%5E2+%3D+0
for yours:
x^2 + xy - 2y^2 = 0
(x+2y)(x-y) = 0
slope of first = -1/2
slope of 2nd = 1
tanθ = (1 + 1/2)/(1 + (1)(-1/2)
= (3/2)/(1/2)
= 3
θ = 71.56 degrees, not 45
see: https://www.wolframalpha.com/input/?i=plot+x%5E2+%2B+xy+-+2y%5E2+%3D+0
So for x^2 + xy - ky^2 = 0
we need
(x+ay)(x+by) = 0
where a+b=1 and ab = -k
also the two slopes are -1/a and -1/b
tanθ = (-1/a + 1/b)/(1 + (-1/a)(-1/b))
= (a-b)/(ab) / (1 + 1/ab)
= (a-b)/(ab) / (ab + 1)(ab)
= (a-b)/(ab+1)
but tan45 = 1
(a-b)/(ab+1) = 1
a-b = ab + 1 , but we know that a+b=1 or
b = 1-a
a-(1-a) = a(1-a) + 1
a-1+a = a-a^2 +1
a^2 + a - 2 = 0
(a-2)(a+1) = 0
a = 2 or a = -1 , then
b = -1 or b = 2
so we could have
(x+2y)(x-y) = 0 ---> x^2 + xy - 2y^2) = 0 , OUR CASE
OR
(x+y)(x-2y) = 0 ----> x^2 - xy - 2y^2 = 0, NOT OUR CASE
k = ab = -2
so k = -2 for each case
To have this equation represented by two straight lines, we must have it in a factored form such as
(x + ay)(x - by) = 0
so that a+b = 1 and ab=-k
e.g. if we had
x^2 + xy - 6y^2 = 0
(x+3y)(x-2y) = 0
our two lines would be
x+3y = 0 or x-2y = 0
slope of first = -1/3
slope of 2nd = 1/2
for acute angle formed:
using the formula
tanθ = (m2 - m1)/(1 + m2m1)
tanθ = (1/2 + 1/3)/(1 + (1/2)(-1/3)
= (5/6) / (5/6)
= 1
θ = 45°
What a lucky guess!
check this graph:
https://www.wolframalpha.com/input/?i=plot+x%5E2+%2B+xy+-+6y%5E2+%3D+0
for yours:
x^2 + xy - 2y^2 = 0
(x+2y)(x-y) = 0
slope of first = -1/2
slope of 2nd = 1
tanθ = (1 + 1/2)/(1 + (1)(-1/2)
= (3/2)/(1/2)
= 3
θ = 71.56 degrees, not 45
see: https://www.wolframalpha.com/input/?i=plot+x%5E2+%2B+xy+-+2y%5E2+%3D+0
So for x^2 + xy - ky^2 = 0
we need
(x+ay)(x+by) = 0
where a+b=1 and ab = -k
also the two slopes are -1/a and -1/b
tanθ = (-1/a + 1/b)/(1 + (-1/a)(-1/b))
= (a-b)/(ab) / (1 + 1/ab)
= (a-b)/(ab) / (ab + 1)(ab)
= (a-b)/(ab+1)
but tan45 = 1
(a-b)/(ab+1) = 1
a-b = ab + 1 , but we know that a+b=1 or
b = 1-a
a-(1-a) = a(1-a) + 1
a-1+a = a-a^2 +1
a^2 + a - 2 = 0
(a-2)(a+1) = 0
a = 2 or a = -1 , then
b = -1 or b = 2
so we could have
(x+2y)(x-y) = 0 ---> x^2 + xy - 2y^2) = 0 , OUR CASE
OR
(x+y)(x-2y) = 0 ----> x^2 - xy - 2y^2 = 0, NOT OUR CASE
k = ab = -2
so k = -2 for each case
Answered by
Reiny
I must have made some kind of error in this big mess.
I can't reconcile the fact that k = 6 will produce the 45 degree angle, but both of our solutions somehow showed that k = 2
I will look at it again later on today
I can't reconcile the fact that k = 6 will produce the 45 degree angle, but both of our solutions somehow showed that k = 2
I will look at it again later on today
Answered by
clive
But I did it comparing with ax^2 + 2hxy + by^2=0 and got -2
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.