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If t1 and ...

Question

If $t_{1}$ and $t_{2}$ are the roots of the equation $x^{2} - 4 x + 2 = 0$ then the point of intersection of tangents at $t_{1}$ and $t_{2}$ on $x y = c^{2}$ is:

(\frac{c}{2}, 2 c)

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(2 c, \frac{c}{2})

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(\frac{c}{2}, c)

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(c, \frac{c}{2})

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Solution

The correct option is D $(c, \frac{c}{2})$
Tangent of $x y =$ $c^{2} a t^{'} t_{1}^{'} a n d^{'} t_{2}^{'}$ are
$x = t_{1}^{2} y - 2 c t_{1}$ .....(i)
and $x = t_{2}^{2} y - 2 c t_{2}$ .....(ii)
On solving (i) and (ii) we get
$y = \frac{2 c}{t_{1} + t_{2}} = \frac{2 c}{4}, x = \frac{2 c t_{1} t_{2}}{t_{1} + t_{2}} = \frac{4 c}{4}$
$∴$ point of intersection is $(c, \frac{c}{2})$
Hence, option 'D' is correct.

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=

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