Q.Prove that the product of lengths of perpendiculars drawn from $P(x_1, y_1)$ to the lines represented by $ax^2 + 2hxy + by^2 = 0$ is $\dfrac{|ax_1^2 + 2hx_1y_1 + by_1^2|}{\sqrt{(a-b)^2+4h^2}}$.

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

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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 $=\dfrac ab=\dfrac{1}{-3}=-\dfrac13$

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

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

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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 $x$:

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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 $y^2$: $d=1$ …