Find the condition such that one of the lines defined by $ax^2+2hxy+by^2=0$ has slope $k$ times that of the other

Question

Find the condition that the lines represented by $$ax^2+2hxy+by^2=0$$ are such that the slope of one line is $k$ times that of the other.

I calculated the two represented by $ax^2+2hxy+by^2=0$ by multiplying by a on both sides and adding $h^2y^2$ to both sides as : $(ax+by)^2=y^2(h^2-ab) (ax+by)=\pm\sqrt{h^2-ab}$. After this I could not understand what the question says.

Answer 1

Let the equations of the straight lines be $$y=mx, y=kmx$$

$$\iff0=(mx-y)(kmx-y)\iff km^2x-mxy(k+1)+y^2=0\ \ \ \ (1)$$

This will be same as $$ax^2+2hxy+by^2=0\ \ \ \ (2)$$

iff $$\dfrac a{km^2}=\dfrac{2h}{-m(k+1)}=\dfrac b1$$

$$\implies m^2=\dfrac a{kb}, m=-\dfrac{2h}{b(k+1)}$$

Eliminate $m$ using $$m^2=(m)^2$$

Answer 2

Dividing the whole equation by x, $ax^2+2hxy+by^2=0$ changes to $$b(\frac{y}{x})^2+2h\frac{y}{x}+a=0$$ This is a quadratic in $\frac{y}{x}$ and also note that $\frac{y}{x}$ represents the slope of the line whose solution is this equation, so letting $\frac{y}{x}$=t, We get $$bt^2+2ht+a=0$$ according to given conditions, slope of one line is 'k' times the other, let slope of one line be 'm' $\implies$slope of other line is 'km', as these are the roots of this equation, we can apply the product and sum conditions for the quadratic and eliminate 'm', yielding the required condition as $$\frac{4h^2k}{b(1+k)^2}=a$$ also note that when k=-1, we have h=0, (by the original equation, sum of roots is 0) so, for k=-1, we can apply the limit $k->-1 $ and $h->0$

Answer 3

Hint Since the lines both pass through $(0, 0)$, we may write them as $$y = m x \quad \textrm{and} \quad y = (km) x,$$ and so the equation of the pair of lines is $$(y - m x)(y - k m x) = 0,$$ or, expanding, $$k m^2 x^2 + [-m(k + 1) x y] + y^2 = 0.$$ We can freely multiply this by a nonzero constant $b$: $$b k m^2 x^2 + b[-m(k + 1) x y ] + by^2 = 0 .$$ So, an equation $$a x^2 + 2 h x y + b y^2 = 0$$ has this form iff there are constants $k, m$ such that $b k m^2 = a$ and $-mb(k + 1) = 2 h$.

To eliminate $m$ (and hence produce a condition in $a, h, b, k$ alone), we square both sides of the second equation and multiply by $k$, giving $$4 h^2 k = -b^2 m^2 (k + 1)^2 k .$$ Substituting using the first equation gives $$\boxed{4 h^2 k = a b (k + 1)^2} .$$ Of course, squaring is not a reversible operation, but this poses no problem for us here, as for our purposes it simply discards the sign of $m$, so the boxed equation is a sufficient and necessary condition.