ParabolaHardBloom L3
Question
$Px + 2y = 1$ is normal to parabola $y^{2} = 4ax$ for :
Options
A.no value of $P$
B.exactly one value of $P$
C.exactly two values of $P$
D.exactly three values of $P$
Solution
Sol. [Key Concept: A line is a normal only in the form $y = - tx + 2at + {at}^{3}$; substituting the given point gives a cubic in $t$ — count its real roots.]
Equation of normal to $y^{2} = 4ax$
$$y = - tx + 2at + at^{3} $$Put $\left( 0,\frac{1}{2} \right)$
$$\begin{matrix} & 2at + at^{3} = \frac{1}{2} \\ & f(t) = 2at^{3} + 4at - 1 \\ & f^{'}(t) = 6at^{2} + 4a = 2a\left( 3t^{2} + 2 \right) \neq 0 \end{matrix}$$
$\Rightarrow f(t)$ can have only one real root.
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