Find the equation to the straight line joining the points $(a t_{1}, \frac{a}{t_{1}})$ and $(a t_{2}, \frac{a}{t_{2}})$ .

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$x + y t_{1} t_{2} - a (t_{1} + t_{2}) = 0$

$x + y t_{1} t_{2} + a (t_{1} t_{2}) = 0$

$\Rightarrow x + y (t_{1} + t_{2}) - a t_{1} t_{2} = 0$

None of these

answer is A.

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Detailed Solution

Given to find the equation of the line joining the points $(a t_{1}, \frac{a}{t_{1}})$ and $(a t_{2}, \frac{a}{t_{2}})$ .

The formula of slope of a line with given two points $(x, y), (x', y')$ through which line is passing is $m = \frac{y' - y}{x' - x}$ .

The equation of a line with slope $m$ and passing through the point $(x', y')$ is $y - y' = m (x - x') .$

Thus, the slope of lines passing through given points is $m = \frac{\frac{a}{t_{2}} - \frac{a}{t_{1}}}{a t_{2} - a t_{1}} = \frac{t_{1} - t_{2}}{t_{1} t_{2} (t_{2} - t_{1})} = \frac{- 1}{t_{1} t_{2}}$ .

The equation of the line passing through the given points is,

$\Rightarrow y - \frac{a}{t_{2}} = \frac{- 1}{t_{1} t_{2}} (x - a t_{2}) \Rightarrow t_{1} (y t_{2} - a) = - x + a t_{2} \Rightarrow x + y t_{1} t_{2} - a (t_{1} + t_{2}) = 0$