ParabolaHardBloom L3
Question
The locus of the foot of the perpendiculars drawn from the vertex on a variable tangent to the parabola $y^{2} = 4ax$ is :
Options
A.$x\left( x^{2} + y^{2} \right) + ay^{2} = 0$
B.$y\left( x^{2} + y^{2} \right) + ax^{2} = 0$
C.$x\left( x^{2} - y^{2} \right) + ay^{2} = 0$
D.None of these
Solution
Sol. [Key Concept: Foot of the perpendicular from the vertex to a variable tangent: eliminate $t$ from the tangent and the perpendicular through $O$ to get $x(x^{2} + y^{2}) + {ay}^{2} = 0$.]
Equation of tangent ' $L$ ' is
$$\begin{array}{r} yt - x = at^{2}\#(1) \end{array}$$
Also $L$ is given by $hx + ky = h^{2} + k^{2}$
(1) and (2) are identical
$${\Rightarrow \frac{t}{k} = - \frac{1}{h} = \frac{at^{2}}{h^{2} + k^{2}} \Rightarrow t = - \frac{k}{h} }{\therefore\ at = \frac{h^{2} + k^{2}}{k} = a\left( - \frac{k}{h} \right) }{\Rightarrow x\left( x^{2} + y^{2} \right) + ay^{2} = 0 }$$
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