$\begin{array}{1 1}(a)\;h+a=0\\(b)\;h+2a=0\\(c)\;h=a\\(d)\;\text{None of these}\end{array}$

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Equation of tangents at point $'t_1'$ and $'t_2'$ are

$t_1y=x+at_1^2$------(1)

$t_2y=x+at_2^2$------(2)

$h=at_1t_2$

$k=a(t_1+t_2)$

Slope of tangents is $\large\frac{1}{t_1}$ & $\large\frac{1}{t_2}$ respectively.

Hence $\large\frac{1}{t_1}\times \large\frac{1}{t_2} $$=-1

$t_1t_2=-1$

Now $h=at_1t_2$

$t_1t_2=-1$

$\Rightarrow h+a=0$

Hence (a) is the correct answer.

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