91Ó°ÊÓ

Let's consider a parabola defined by the equation \(y^2=4ax\) with its axis parallel to the x-axis. The focus of this parabola is the point \(F(a,0)\) and its directrix is the vertical line \(x=-a\).

Let \(P\) be a point on the parabola, say \(P(at^2, 2at)\). By the rule of tangents for a parabola, the tangent to the parabola at point \(P\) can be expressed as \(ty=x+at^2\).

Let \(Q\) be the point on directrix \(x=-a\) and the tangent to the parabola intersects the directrix at this point. As \(Q\) is on line \(x=-a\) and also on the tangent line, we get: \(t(-a)+at^2=-a\). Solving for \(Q\), we obtain the coordinates \(Q(-a, -2a(t-1))\).

We need to show that \(\angle FPQ = 90^\circ\). To do this, we'll find the slopes of lines \(FP\) and \(FQ\) and show that they are negative reciprocals. The slope of the line \(FP\) is: $$m_{FP}= \frac{2at-0}{at^2-a} = \frac{2t}{t^2-1} $$ The slope of the line \(FQ\) is: $$m_{FQ} = \frac{-2a(t-1)-0}{-a-a} = \frac{2(t-1)}{-2} =\frac{1-t}{1} $$ The product of the slopes of the lines \(FP\) and \(FQ\) is: $$m_{FP} \cdot m_{FQ} = \frac{2t}{t^2-1} \cdot \frac{1-t}{1} = -1 $$ Since the product of the slopes is -1, the two lines \(FP\) and \(FQ\) are perpendicular to each other, which means \(\angle FPQ = 90^\circ\). Hence, it is proved that a portion of a tangent to a parabola intercepted between the directrix and the curve subtends a right angle at the focus.