Geo5.1.27

Find the locus of the midpoints of chords of the parabola $y^2=4ax\,$ which subtend a right angle at the vertex of the parabola.

Let $P(x_1,y_1)\,$ be the midpoint of the chord BC of the parabola $y^2=4ax\,$

Equation of BC is $yy_1-2a(x+x_1)=y_1^{2}-4ax_1\,$

$yy_1-2ax=y_1^{2}-2ax_1\,$

Join AB,AC. Given angle BAC is a right angle,homogenising $y^2=4ax\,$

$y^2-4ax[\frac{yy_1-2ax}{y_1^2-2ax}]=0\,$

$(y_1^{2}-2ax_1)y^2-4ay_1 xy+8a^2x^2=0\,$

This is the combined equation of AB and AC. $\angle BAC=90^\circ\,$

coefficient of $x^2\,$ + coefficient of $y^2\,$ =0

$8a^2+(y_1^{2}-2ax_1)=0\,$

$y_1^{2}=2a(x_1-4a)\,$

Therefore the locus of P is

$y^2=2a(x-4a)\,$