Question

The locus of points such that two of the three normals drawn from them to the parabola $y^2=4ax$ coincide is

• A. $27a{y}^2=4(x-2a)$
• B. $27a{y}^2=4(x-2a{)}^3$
• C. $27ay=4(x-2a{)}^2$
• D. $27ay=4(x-2a)$

Answer 1

Answer: (B)

$27a{y}^2=4(x-2a{)}^3$

Equation of normal to parabola is
$y=2at+a{t}^3-tx$
As it should have a repeated root , $y^{\prime }=0$ should have a single root
$\Rightarrow y^{\prime }=2a-x+3a{t}^2=0$
$\Rightarrow 3a{t}^2=x-2a$
Substituting $t$ in $e{q}^n$, we get
$y=(2a-x)t+a{t}^3$
$y=(2a-x)\sqrt{\frac{(x-2a)}{3a}}+a\sqrt{{\left(\frac{x-2a}{3a}\right)}^3}$
$3\sqrt{3}ay=(3-1)(x-2a{)}^{3/2}$
Squaring on both sides,
$27a{y}^2=4(x-2a{)}^3$