Question

Locus of the point of intersection of tangents at the end points of normal chord of the parabola $y^2=4ax$ is

• A. $x{y}^2+2a(4a^2+y^2)=0$
• B. $x{y}^2=2a(2a^2+y^2)$
• C. $x{y}^2+a(2a^2+y^2)=0$
• D. $x{y}^2=a(2a^2+y^2)$

Answer 1

The correct option is A $x{y}^2+2a(4a^2+y^2)=0$

Let point of intersection be $(x_1,y_1)$
Then normal chord is the chord of contact

$\Rightarrow y{y}_1=2a(x+x_1)$

$y+t_{1}x=2a{t}_1+2at{{}_1}^3$ is equation of normal chord

$\Rightarrow \frac{y_1}{1}=\frac{-2a}{t_1}=\frac{2a{x}_1}{2a{t}_1+2at{{}_1}^3}$

$y_1=\frac{-2a}{t_1}$ and $x_1=-2a(1+t{{}_1}^2)$

$t_1=\frac{-2a}{y_1}$

$\Rightarrow x_1=-2a(1+\frac{4a^2}{y{{}_1}^2})$

$\Rightarrow xy{{}_1}^2=-2a(y{{}_1}^2+4a^2)$

$\therefore$ required locus is $x{y}^2+2a(y^2+4a^2)=0$