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Question

A variable line, drawn through the point of intersection of the straight lines $$\cfrac{x}{a}+\cfrac{y}{b}=1$$ and $$\cfrac{x}{b}+\cfrac{y}{a}=1$$, meets the coordinate axes in $$A$$ and $$B$$. Show that the locus of the mid point of $$AB$$ is the curve $$2xy(a+b)=ab(x+y)$$


Solution

$$\dfrac{x}{a}+\dfrac{y}{b}=1$$
$$\Rightarrow bx+ay-ab=0$$         ...(i)
$$\dfrac{x}{b}+\dfrac{y}{a}=1$$
$$\Rightarrow ax+by-ab=0$$         ...(ii)

(i) $$\times a$$        $$abx+a^2y-a^2b=0$$
(ii) $$\times b$$       $$\underline{abx+b^2y-ab^2=0}$$
              $$(a^2-b^2)y=a^2b-ab^2$$

$$\Rightarrow (a-b)(a+b)y=ab(a-b)$$
$$\Rightarrow y=\dfrac{ab}{a+b}$$

$$\therefore x=\dfrac{ab}{a+b}$$

$$y-\dfrac{ab}{a+b}=m\left(x-\dfrac{ab}{a+b}\right)$$

$$\therefore A\left(\dfrac{ab}{a+b}\dfrac{(m-1)}{m}, 0\right), B\left(0, \dfrac{ab}{a+b}(1-m)\right)$$

$$\therefore A\left(\dfrac{ab}{2(a+b)} x\dfrac{(m-1)}{m}, \dfrac{ab}{2(a+b)} x(1-m)\right)$$

$$x=\dfrac{ab(m-1)}{2(a+b)}, y=\dfrac{ab(1-m)}{2(a+b)}$$
Equation $$'m'$$
$$2xy(a+b)=ab(x+y)$$ (proved)

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