ParabolaHardBloom L3
Question
If the normal to a parabola $y^{2} = 4ax$ at $P$ meets the curve again in $Q$ and if $PQ$ and the normal at $Q$ makes angles $\alpha$ and $\beta$ respectively with the $x$-axis then $tan\alpha(tan\alpha + tan\beta)$ has the value equal to :
Options
A.0
B.-2
C.$- 1/2$
D.-1
Solution
Sol. [Key Concept: With the normal at $P$ of slope $\tan\alpha = - t$ meeting again at $Q$, the relation $t_{2} = - t - \frac{2}{t}$ yields $\tan\alpha(tan\alpha + tan\beta) = - 2$.]
$$t_{2} = - t_{1} - \frac{2}{t_{1}}$$
$$\begin{matrix} & \left( t_{1} + t_{2} \right)t_{1} = - 2 \\ & \ - t_{1} = tan\alpha, - t_{2} = tan\beta \end{matrix}$$
$$\Rightarrow tan\alpha(tan\alpha + tan\beta) = - 2 $$
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