Question

Hence find the locus of the orthocentre of a triangle of which two sides are given in position and whose third side goes through a fixed point.

Answer 1

Let the line $lx+my=1$ passes through some fixed point, $P\equiv (\alpha ,\beta )$ then $l\alpha +m\beta =\mathrm{1....1}$$(x^{{}^{\prime }},y^{{}^{\prime }})$
If $\frac{x^{{}^{\prime }}}{l}=\frac{y^{{}^{\prime }}}{m}=\frac{a+b}{a{m}^2-2hlm+b{l}^2}=\frac{x^{{}^{\prime }}+y^{{}^{\prime }}}{l+m}....2$ be the other center
$\frac{x^{{}^{\prime }}}{l}=\frac{y^{{}^{\prime }}}{m}=\frac{\alpha x^{{}^{\prime }}}{\alpha l}=\frac{\beta y^{{}^{\prime }}}{\beta m}=\frac{\alpha x^{{}^{\prime }}+\beta y^{{}^{\prime }}}{l\alpha +\beta m}$
As $=\alpha x^{{}^{\prime }}+\beta y^{{}^{\prime }}[\because l\alpha +m\beta =1]$
$\frac{x^{{}^{\prime }}}{l}=\frac{y^{{}^{\prime }}}{m}=\alpha x^{{}^{\prime }}+\beta y^{{}^{\prime }}$
$\because l=\frac{x^{{}^{\prime }}}{\alpha x^{{}^{\prime }}+\beta y^{{}^{\prime }}},m=\frac{y^{{}^{\prime }}}{\alpha x^{{}^{\prime }}+\beta y^{{}^{\prime }}}$
Putting these values in $2$ we get
$\frac{a+b}{a{\left(\frac{y^{{}^{\prime }}}{\alpha x^{{}^{\prime }}+\beta y^{{}^{\prime }}}\right)}^2-2h\frac{x^{{}^{\prime }}}{\alpha x^{{}^{\prime }}+\beta y^{{}^{\prime }}},\frac{y^{{}^{\prime }}}{\alpha x^{{}^{\prime }}+\beta y^{{}^{\prime }}}+b{\left(\frac{x^{{}^{\prime }}}{\alpha x^{{}^{\prime }}+\beta y^{{}^{\prime }}}\right)}^2}=\frac{x^{{}^{\prime }}+y^{{}^{\prime }}}{\frac{x^{{}^{\prime }}}{\alpha x^{{}^{\prime }}+\beta y^{{}^{\prime }}}+=\frac{y^{{}^{\prime }}}{\alpha x^{{}^{\prime }}+\beta y^{{}^{\prime }}}}$
or $\frac{(a+b){(\alpha x^{{}^{\prime }}+\beta y^{{}^{\prime }})}^2}{a{y}^{{{}^{\prime }}^2}-x^{{}^{\prime }}{y}^{{}^{\prime }}+b{x}^{{}^{\prime }{{}^{\prime }}^2}}=\frac{(x^{{}^{\prime }}+y^{{}^{\prime }})(\alpha x^{{}^{\prime }}+\beta y^{{}^{\prime }})}{x^{{}^{\prime }}+y^{{}^{\prime }}}$
Hence the required locus is
$(a+b)(\alpha x+\beta y)=a{x}^2-2hxy+b{x}^2$