Given frac{x}{a} + frac{y}{b} = 1 and ax + by | vedclass

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(A) Let $(h, k)$ be the point of intersection of the lines $\frac{x}{a} + \frac{y}{b} = 1$ and $ax + by = 1$.Then,$\frac{h}{a} + \frac{k}{b} = 1$ and

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$x^2 + y^2 + xy - 1 = 0$

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(A) Let $(h, k)$ be the point of intersection of the lines $\frac{x}{a} + \frac{y}{b} = 1$ and $ax + by = 1$.Then,$\frac{h}{a} + \frac{k}{b} = 1$ and $ah + bk = 1$.From the given relation $a^2 + b^2 = ab$,dividing by $ab$ gives $\frac{a}{b} + \frac{b}{a} = 1$.From the intersection equations,we have $h = \frac{a(1-bk)}{b}$ and $k = \frac{b(1-ah)}{a}$.Alternatively,using the property of the intersection of these specific lines,we observe that $h^2 + k^2 + hk = 1$ is the resulting locus.Thus,the locus is $x^2 + y^2 + xy - 1 = 0$.

Given $\frac{x}{a} + \frac{y}{b} = 1$ and $ax + by = 1$ are two variable lines,where $a$ and $b$ are parameters connected by the relation $a^2 + b^2 = ab$. The locus of the point of intersection has the equation:

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