Show that the locus is the directrix.

Question

P and Q are points on the parabola x=2at, y =at^2.  The tangents at P and Q intersecdt at R so that the anbgle PQR=90.  Show that the locus of R is the directrix.

Answer 1

We can find how y and x are related by substituting t=x/(2a):

y=a(x²/(4a²)=x²/(4a), a parabola with vertex (0,0).

On the axis of symmetry (y-axis) the focus and directrix line are equidistant from the vertex. Call this distance f. All points P(x,y) on the parabola are equidistant from the focus F(0,f) and directrix line (y=-f).

PF²=(f-y)²+x²=f²-2fy+y²+x² 

The y-coord of P is distance y-(-f)=y+f from the directrix line y=-f.

Since x²=4ay we can substitute:

PF=√(f²-2fy+y²+4ay).

Therefore:

√(f²-2fy+y²+4ay)=y+f,

f²-2fy+y²+4ay=y²+2fy+f²,

-2fy+4ay=2fy,

4ay=4fy⇒f=a.

The tangent at P(x_P,y_P) is dy/dx=x_P/(2a) and at Q(x_Q,y_Q) dy/dx=x_Q/(2a).

The tangent lines are respectively:

y-y_P=(x_P/(2a))(x-x_P) and y-y_Q=(x_Q/(2a))(x-x_Q).

Their intersection is given by:

(x_P/(2a))(x-x_P)+y_P=(x_Q/(2a))(x-x_Q)+y_Q.

(x_P/(2a))(x-x_P)-(x_Q/(2a))(x-x_Q)=y_Q-y_P,

x(x_P-x_Q)=2a(y_Q-y_P)+x_P²-x_Q²,

x=(2a(y_Q-y_P)+x_P²-x_Q²)/(x_P-x_Q).

y_P=x_P²/(4a), y_Q=x_Q²/(4a), 2a(y_Q-y_P)=½(x_Q²-x_P²), 

x=½(x_P²-x_Q²)/(x_P-x_Q)=½(x_P+x_Q);

y=(x_P/(2a))(x-x_P)+y_P,

y=(x_P/(2a))(½(x_P+x_Q)-x_P)+y_P,

y=(x_P/(4a))(x_Q-x_P)+y_P=(x_P/(4a))(x_Q-x_P)+x_P²/(4a),

y=x_Px_Q/(4a).

This makes the intersection point X of the tangents at P and Q X(½(x_P+x_Q),x_Px_Q/(4a)).

More to follow in due course...