Question

Two fixed points $A$ and $B$ are taken on the axes such that $OA=a$ and $OB=b$; two variable points $A^{{}^{\prime }}$ and $B^{{}^{\prime }}$ are taken on the same axes; find the locus of the intersection of $A{B}^{{}^{\prime }}$ and $A^{{}^{\prime }}B$
(1) when $O{A}^{{}^{\prime }}+O{B}^{{}^{\prime }}=OA+OB$,
and (2) when $\frac{1}{O{A}^{{}^{\prime }}}-\frac{1}{O{B}^{{}^{\prime }}}=\frac{1}{OA}-\frac{1}{OB}$.

Answer 1

$A(a,0)$ and $B(0,b)$

Let $A^{\prime }$ be $(a^{\prime },0)$ and $B^{\prime }$ be $(0,b^{\prime })$ and point of intersection be $P(h,k)$

Equation of $A{B}^{\prime }$ is

$y-0=\frac{b^{\prime }-0}{0-a}(x-a)ay+b^{\prime }x=a{b}^{\prime }$

It passes through $(h,k)$

$\Rightarrow ak+b^{\prime }h=a{b}^{\prime }ak=a{b}^{\prime }-b^{\prime }h\Rightarrow b^{\prime }=\frac{ak}{a-h}.......\left(i\right)$

Equation of $A^{\prime }B$ is

$y-0=\frac{b-0}{0-a^{\prime }}(x-a^{\prime })bx+a^{\prime }y=a^{\prime }b$

It also passes through $(h,k)$

$\Rightarrow bh+a^{\prime }k=a^{\prime }bbh=a^{\prime }(b-k)a^{\prime }=\frac{bh}{(b-k)}......\left(ii\right)$

(1) Given $O{A}^{\prime }+O{B}^{\prime }=OA+OB$

$a^{\prime }+b^{\prime }=a+b$

using $\left(i\right)$ and $\left(ii\right)$

$\frac{bh}{b-k}+\frac{ak}{a-h}=a+b\frac{abh-b{h}^2+abk-a{k}^2}{(b-k)(a-h)}=a+b(a+b)(b-k)(a-h)=abh-b{h}^2+abk-a{k}^2(a+b)\{ab-bh-ak+hk)=abh-b{h}^2+abk-a{k}^{2}b{h}^2+a{k}^2+ab(a+b)+(a+b)hk-a(a+b)k-abk-b(a+b)h-abh=0b{h}^2+a{k}^2+(a+b)hk-a(a+2b)k-b(2a+b)h+ab(a+b)=0$

Replacing $h$ by $x$ and $y$ by $k$

$b{x}^2+a{y}^2+(a+b)xy-a(a+2b)y-b(2a+b)x+ab(a+b)=0$

is the required locus

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(2) Given $\frac{1}{O{A}^{\prime }}-\frac{1}{O{B}^{\prime }}=\frac{1}{OA}-\frac{1}{OB}$

$\frac{b-k}{bh}-\frac{a-h}{ak}=\frac{1}{a}-\frac{1}{b}\frac{abk-a{k}^2-abh+b{h}^2}{abhk}=\frac{b-a}{ab}abk-a{k}^2-abh+b{h}^2=bhk-ahka{k}^2-ahk+bhk-b{h}^2+abh-abk=0ak(k-h)+bh(k-h)-ab(k-h)=0(k-h)(ak+bh-ab)=0\Rightarrow k-h=0$

Replacing $h$ by $x$ and $k$ by $y$

$y-x=0$

$y=x$