The point of intersection of the tengents to the parabola $y^{2} = 4 x$ at the points, where the parameter 't' has the value 1 and 2, is

(3, 8)

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(1, 5)

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(2, 3)

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(4, 6)

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Solution

The correct option is C
(2, 3)

Slope of the tanget to a parabola at t is $\frac{1}{t}$ . The equations of tangents at $t_{1} a n d t_{2}$ are
$y - 2 a t_{1} = \frac{1}{t_{1}} (x - a t_{1}^{2}) . . . . . (i) y - 2 a t_{1} = \frac{1}{t_{1}} (x - a t_{1}^{2}) . . . . . (i i)$
Solving (i) and (ii), the point of intersection is $(a t_{1} t_{2}, a (t_{1} + t_{2})) w h e r e t_{1} = 1 a n d t_{2} = 2$ . Here a=1.
Hence the required point is $(2, 3)$

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