Question

Locus of mid points of normal chords of the parabola $y^2=4ax$ is :

• A. $8a^4+4a^2{y}^2+y^2(y^2-4ax)=0$
• B. $12a^4+7a^2{y}^2+y^2(y^2-4ax)=0$
• C. $10a^4+6a^2{y}^2+y^2(y^2-8ax)=0$
• D. $8a^4+9a^2{y}^2+y^2(y^2-7ax)=0$

Answer 1

The correct option is A $8a^4+4a^2{y}^2+y^2(y^2-4ax)=0$

Equation of chord having $P(x_1,y_1)$ as its mid points is
$T=S_1$
$\Rightarrow y{y}_1-2ax=y_1^2-4a{x}_1$
$\Rightarrow y=\frac{2a}{y_1}x+y_1-\frac{4a{x}_1}{y_1}\cdots \left(1\right)$
Equation of normal will be
$y=mx-2am-a{m}^3\cdots \left(2\right)$
$\because$ both the equation represents the same line
$\therefore m=\frac{2a}{y_1}$
and $y_1-\frac{4a{x}_1}{y_1}=-2am-a{m}^3$
$\Rightarrow y_1^2-4a{x}_1=-y_1(\frac{4a^2}{y_1}+\frac{8a^4}{y_1^3})\Rightarrow y_1^2(y_1^2-4a{x}_1)+4a^2{y}_1^2+8a^4=0$
Locus of $P$ is $8a^4+4a^2{y}^2+y^2(y^2-4ax)=0$