If the three normals drawn to the parabola y^2=2x pass through the point (a,0), a0, then a…

Question

If the three normals drawn to the parabola $y^2=2x$ pass through the point $(a,0)$, $a
e0$, then $a$ must be greater than

Accepted Answer

Normal to $y^2=4Ax$ ($A=\frac12$): $tx+y=2At+At^3=t+\frac{t^3}2$, i.e. $t^3+(2-2x)t-2y=0$.

Through $(a,0)$: $t^3+(2-2a)t=0$. With roots $t_1,t_2,t_3$: $\sum t_1=0$, $\sum t_1t_2=2-2a$.

Requiring three distinct real roots means $\sum t_i^2>0$:
$$\left(\sum t_i\right)^2-2\sum t_1t_2>0\ \Longrightarrow\ -2(2-2a)>0\ \Longrightarrow\ a>1.$$

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