Question

A straight line $AB$, whose length is $c$, slides between two given oblique axes which meet at $O$; find the locus of the orthocentre of the triangle $OAB$.

Answer 1

Taking $O$ as the origin $,OA$ and $OB$ as the axes$,\angle AOB=\omega$ $,OA=a$ and $OB=b$.

Then equation of $AL$ perpendicular from $A$ to $OB$ is

$x\cos \omega +y=a\cos \omega .....\left(i\right)$

Similarly, equation of $BM$ perpendicular from $B$ on $OA$ is

$x+y\cos \omega =b\cos \omega .....\left(ii\right)$

Given $AB=c$

$c^2=a^2+b^2-2ab\cos \omega$

Substituting $a$ and $b$ from $\left(i\right)$ and $\left(ii\right)$ respectively, we get

$c^2={\left(\frac{x\cos \omega +y}{\cos \omega }\right)}^2+{\left(\frac{x+y\cos \omega }{\cos \omega }\right)}^2-2\left(\frac{x\cos \omega +y}{\cos \omega }\right)\left(\frac{x+y\cos \omega }{\cos \omega }\right)\cos \omega c^2{\cos }^2\omega =x^2{\cos }^2\omega +y^2+2xy\cos \omega +x^2+y^2{\cos }^2\omega +2xy\cos \omega -2\cos \omega \{x^2\cos \omega +xy{\cos }^2\omega +xy+y^2\cos \omega \}{c}^2{\cos }^2\omega =(1-{\cos }^2\omega )x^2+{(1-{\cos }^2\omega )y}^2+2xy\cos \omega -2xy{\cos }^3\omega c^2{\cos }^2\omega =(1-{\cos }^2\omega )x^2+{(1-{\cos }^2\omega )y}^2+2xy\cos \omega (1-{\cos }^2\omega )c^2{\cot }^2\omega =x^2+y^2+2xy\cos \omega x^2+y^2+2xy\cos \omega -c^2{\cot }^2\omega =0$

Clearly the equation represents a circle.

Hence the locus of orthocentre of triangle is a circle.