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Problem:

If the lines joining the origin and the points of intersection of curves $ax^2+2hxy+by^2+2gx=0$ and $a_1x^2+2h_1xy+b_1y^2+2g_1x=0$ are mutually perpendicular, prove that $g(a_1+b_1)=g_1(a+b)$.

$$$$ If the curve intersected a chord, then we would have to make the curve homogeneous with the equation of the chord. I don't know how to extend this to this problem.

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Let the equations of the one of the perpendicular lines be $y=mx$

$\implies$ the equation of the other will be $x+my=0$

So, the equation of the pair straight lines $$(mx-y)(x+my)=0\iff mx^2+xy(m^2-1)-my^2=0\ \ \ \ (1)$$

Now the equation of the two dimensional curve passing through the points of intersection of the given curves

$$ax^2+2hxy+by^2+2gx+K(a_1x^2+2h_1xy+b_1y^2+2g_1x)=0$$

$$\iff (a+a_1K)x^2+2xy(h+h_1K)+y^2(b+b_1K)+2x(g+g_1K)=0\ \ \ \ (2)$$ where $K$ is an arbitrary constant

Comparing the coefficients of $x^2,y^2,x$ of $(1),(2)$

$$\dfrac m{a+a_1K}=\dfrac{-m}{b+b_1K}=\dfrac0{g+g_1K}$$

$\dfrac m{a+a_1K}=\dfrac{-m}{b+b_1K}\implies a+a_1K+b+b_1K=0\iff K=-\dfrac{a+b}{a_1+b_1}$

$\dfrac{-m}{b+b_1K}=\dfrac0{g+g_1K}\implies g+g_1K=\cdots=0\iff K=-\dfrac g{g_1}$

Can you take it from here?

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    $\begingroup$ @BetterWorld, Please find the updated version & revert to me for any confusion. $\endgroup$ – lab bhattacharjee Aug 13 '15 at 16:16

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