Question

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 ?

Answer 1

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

Answer 2

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

Answer 3

But I did it comparing with ax^2 + 2hxy + by^2=0 and got -2