Q.Show that lines represented by $x^2 + 6xy + 9y^2 = 0$ are coincident.

Compare with $ax^2+2hxy+by^2=0$; sum of slopes $=-2h/b$, product $=a/b$.

Compare $x^2+kxy-3y^2=0$ with $ax^2+2hxy+by^2=0$: $a=1,\ 2h=k \Rightarrow h=\dfrac{k}{2},\ b=-3$.

Sum of slopes $=-\dfrac{2h}{b}=-\dfrac{k}{-3}=\dfrac{k}{3}$

Product of slopes $=\dfrac{a}{b}=\dfrac{1}{-3}=-\dfrac13$

Given: sum $=2\times$ product:

$\frac{k}{3}=2\left(-\frac13\right)=-\frac23 \implies k=-2$

Sum of slopes $=\dfrac{-2h}{b}$, product $=\dfrac ab$; set sum $=2\times$product.

For $x^2+kxy-3y^2=0$: $a=1,\ 2h=k,\ b=-3$

Sum of slopes $=\dfrac{-2h}{b}=\dfrac{-k}{-3}=\dfrac k3$

Product of slopes $=\dfrac ab=\dfrac{1}{-3}=-\dfrac13$

Given: sum $=2\times$product: $\dfrac k3=2\left(-\dfrac13\right)=-\dfrac23 \Rightarrow k=-2$

Since $h^2=ab$, the discriminant of the slope-difference is zero, so the slopes are equal.

For $ax^2+2hxy+by^2=0$: $m_1+m_2=\dfrac{-2h}{b}$, $m_1m_2=\dfrac ab$.

$(m_1-m_2)^2=(m_1+m_2)^2-4m_1m_2=\dfrac{4h^2}{b^2}-\dfrac{4a}{b}=\dfrac{4(h^2-ab)}{b^2}$

Since $h^2=ab$, $(m_1-m_2)^2=0 \Rightarrow m_1=m_2$.

Ratio $m_1:m_2=1:1$.

Substitute $y=-2x$ (from $2x+y=0$) into the pair of lines equation; it must vanish identically in $x$.

If $2x+y=0$, i.e. $y=-2x$, is one of the lines represented by $3x^2+kxy+2y^2=0$, substituting $y=-2x$ must give $0$ for all $x$:

$3x^2+kx(-2x)+2(-2x)^2 = 3x^2-2kx^2+8x^2 = (11-2k)x^2 = 0$

For this to hold for all $x$: $11-2k=0 \implies k=\dfrac{11}{2}$

Since $2x+y=0$ is one factor, find the matching second factor of $6x^2+kxy+y^2$ and compare coefficients.

The pair $6x^2+kxy+y^2=0$ factors as $(2x+y)(cx+dy)=0$ for some constants $c,d$ (since $2x+y=0$ is given as one line).

Expanding: $(2x+y)(cx+dy)=2cx^2+(2d+c)xy+dy^2$.

Compare with $6x^2+kxy+y^2$:

• Coefficient of $x^2$: $2c=6 \Rightarrow c=3$
• Coefficient of $y^2$: $d=1$
• Coefficient of $xy$: $k=2d+c=2(1)+3=5$

So the second line is $3x+y=0$, and:

$(2x+y)(3x+y)=6x^2+2xy+3xy+y^2=6x^2+5xy+y^2$

which matches $6x^2+kxy+y^2$ with $k=5$.