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(B) The equation of a line passing through the intersection of the lines $\frac{x}{\alpha} + \frac{y}{\beta} = 1$ and $\frac{x}{\beta} + \frac{y}{\alp

Vedclass · Accepted Answer

$\alpha \beta (x + y) = 2xy(\alpha + \beta)$

Vedclass · Answer

(B) The equation of a line passing through the intersection of the lines $\frac{x}{\alpha} + \frac{y}{\beta} = 1$ and $\frac{x}{\beta} + \frac{y}{\alpha} = 1$ is given by the family of lines equation: $\left( \frac{x}{\alpha} + \frac{y}{\beta} - 1 \right) + \lambda \left( \frac{x}{\beta} + \frac{y}{\alpha} - 1 \right) = 0$. Rearranging this,we get $x \left( \frac{1}{\alpha} + \frac{\lambda}{\beta} \right) + y \left( \frac{1}{\beta} + \frac{\lambda}{\alpha} \right) = 1 + \lambda$. This line meets the axes at $A \left( \frac{1 + \lambda}{\frac{1}{\alpha} + \frac{\lambda}{\beta}}, 0 \right)$ and $B \left( 0, \frac{1 + \lambda}{\frac{1}{\beta} + \frac{\lambda}{\alpha}} \right)$. Let $(h, k)$ be the midpoint of $AB$. Then $h = \frac{1 + \lambda}{2(\frac{1}{\alpha} + \frac{\lambda}{\beta})}$ and $k = \frac{1 + \lambda}{2(\frac{1}{\beta} + \frac{\lambda}{\alpha})}$. From these,$\frac{1}{2h} = \frac{\frac{1}{\alpha} + \frac{\lambda}{\beta}}{1 + \lambda}$ and $\frac{1}{2k} = \frac{\frac{1}{\beta} + \frac{\lambda}{\alpha}}{1 + \lambda}$. Adding these,$\frac{1}{2h} + \frac{1}{2k} = \frac{\frac{1}{\alpha} + \frac{1}{\beta} + \lambda(\frac{1}{\alpha} + \frac{1}{\beta})}{1 + \lambda} = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta}$. Thus,$\frac{h + k}{2hk} = \frac{\alpha + \beta}{\alpha \beta}$,which simplifies to $\alpha \beta (h + k) = 2hk(\alpha + \beta)$. The locus of $(h, k)$ is $\alpha \beta (x + y) = 2xy(\alpha + \beta)$.

If a variable line drawn through the point of intersection of straight lines $\frac{x}{\alpha} + \frac{y}{\beta} = 1$ and $\frac{x}{\beta} + \frac{y}{\alpha} = 1$ meets the coordinate axes in $A$ and $B$,then the locus of the midpoint of $AB$ is

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