The three distinct $A (a {t_{1}}^{2}, 2 a t_{1}), B (a {t_{2}}^{2}, 2 a t_{2})$ and $C (a, 0)$ (where $a$ is a real number) are collinear. If

$t_{1} t_{2} = - 1$

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$t_{1} t_{2} = 1$

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$2 t_{1} t_{2} = t_{1} + t_{2}$

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$t_{1} + t_{2} = a$

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Solution

The correct option is A
$t_{1} t_{2} = - 1$

Finding the condition for collinearity:
Given three distinct points $A (a {t_{1}}^{2}, 2 a t_{1}), B (a {t_{2}}^{2}, 2 a t_{2})$ and $C (a, 0)$ .
If the given points are co-linear then the area of triangle made by these points will be zero.
Since the area of the triangle is given by,
$\frac{1}{2} [y_{1} (x_{2} - x_{3}) + y_{2} (x_{3} - x_{1}) + y_{3} (x_{1} - x_{2})] = \frac{1}{2} [2 a t_{1} (a {t_{2}}^{2} - a) + 2 a t_{2} (a - a {t_{1}}^{2}) + 0] = \frac{1}{2} [2 a^{2} t_{1} {t_{2}}^{2} - 2 a^{2} t_{1} + 2 a^{2} t_{2} - 2 a^{2} {t_{1}}^{2} t_{2}] = a^{2} [t_{1} t_{2} (t_{2} - t_{1}) + (t_{2} - t_{1})]$
The area is equal to zero.
Thus, $a^{2} [t_{1} t_{2} (t_{2} - t_{1}) + (t_{2} - t_{1})] = 0$
$\Rightarrow$ $t_{1} t_{2} = - 1$
Hence option (A) is the correct option.

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