Lines joining origin to points of intersection of two conics

Question

If the lines joining origin and point 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, then prove that $$g(a_1+b_1)=g_1(a+b)$$ How do we do this? Do we have to homogenize the $2$nd curve?

Answer 1

The equation of any 2-D curve passing through the intersection of the given two curves can be expressed as $$ax^2+2hxy+by^2+2gx+\lambda(a_1x^2+2h_1xy+b_1y^2+2g_1x)=0$$

$$\iff x^2(a+\lambda a_1)+2xy(h+\lambda h_1)+y^2(b+\lambda b_1)+2x(g+\lambda g_1)=0\ \ \ \ (1)$$ where $\lambda$ is an arbitrary constant

Now the equation of the perpendicular lines through the origin can expressed as $$y=mx, y=-\frac1mx\iff x+my=0$$

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

This is a special case of $(1)$

$$\implies\frac m{a+\lambda a_1}=\frac{m^2-1}{2(h+\lambda h_1)}=\frac{-m}{b+\lambda b_1}=\frac0{g+\lambda g_1}$$

$$\text{Now, }\frac{-m}{b+\lambda b_1}=\frac0{g+\lambda g_1}\text{ and }\frac{m^2-1}{2(h+\lambda h_1)}=\frac0{g+\lambda g_1}\implies g+\lambda g_1=0\iff\lambda=?$$

$$\text{Again, }\frac m{a+\lambda a_1}=\frac{-m}{b+\lambda b_1}\implies a+\lambda a_1=-(b+\lambda b_1)\iff\lambda=?$$

Equate the values of $\lambda$