Question

Locus of the feet of the perpendiculars drawn from the vertex of the parabola $y^2=4ax$ upon all such chords of the parabola which subten a right angle at the vertex is

• A. $x^2+y^2-4ax=0$
• B. $x^2+y^2-2ax=0$
• C. $x^2+y^2+2ax=0$
• D. $x^2+y^2+4ax=0$

Answer 1

The correct option is A $x^2+y^2-4ax=0$

Let the points where line $y=mx+c$ cut parabola be

$\Rightarrow$ $A(x_1,y_1)$ and $B(X_2,y_2)$

$\Rightarrow$ $y^2=4ax$

$\Rightarrow$ $(mx+c{)}^2=4ax$

$\Rightarrow$ $m^2{x}^2+x(2mc-4a)x+c^2=0$

$\Rightarrow$ $x_1{x}_2=\frac{c^2}{m^2}$ and

$\Rightarrow$ $y_1{y}_2=\frac{4ac}{m}$

As $OA$ and $OB$ are perpendicular. $O$ is origin

$\frac{y_1{y}_2}{x_1{x}_2}=-1$

$\Rightarrow$ $\frac{\frac{4ac}{m}}{\frac{c^2}{m^2}}=-1$

$\Rightarrow$ $\frac{4am}{c}=-1$

$\therefore$ $c=-4am$

So line is $y=mx-4am$

Let the foot of perpendicular from origin on this line be $(h,k)$

Slope of $OC=\frac{-1}{Slpoeofline}=\frac{-1}{m}$

$\Rightarrow$ $\frac{k}{h}=\frac{-1}{m}$

$\Rightarrow$ $m=\frac{h}{k}$

Also $(h,k)$ lie on $y=mx-4am$

$\Rightarrow$ $k=mh-4am$

$\Rightarrow$ $k=\frac{-h}{k}\times h-4a\times \frac{-h}{k}$

$\Rightarrow$ $h^2+k^2-4ah=0$

$\therefore$ Locus is $x^2+y^2-4ax=0$