Question

Through the vertex O of the parabola $y^2=4ax$, a perpendicular is drawn to any tangent meeting it at P & the parabola at Q. Show that OP$\cdot$OQ = constant.

Answer 1

$OP \cdot OQ = 4a^2$

Answer 2

The parametric coordinates of a point T on the parabola $y^2 = 4ax$ are $(at^2, 2at)$. The equation of the tangent at T is $yt = a(x+at^2)$, which can be rewritten as $x - yt + at^2 = 0$.

The line OP is perpendicular to the tangent and passes through the vertex O(0,0). The slope of the tangent is $m_t = 1/t$. The slope of OP is $m_{OP} = -t$. The equation of the line OP is $y = -tx$.

P is the foot of the perpendicular from O(0,0) to the tangent line $x - yt + at^2 = 0$. The distance OP is: $OP = \frac{|a t^2|}{\sqrt{1^2 + (-t)^2}} = \frac{|a|t^2}{\sqrt{1+t^2}}$

Q is the point where the line OP ($y = -tx$) intersects the parabola $y^2 = 4ax$. Substituting $y = -tx$ into the parabola equation: $(-tx)^2 = 4ax \implies t^2x^2 = 4ax$ $x(t^2x - 4a) = 0$ The non-vertex solution is $x_Q = \frac{4a}{t^2}$. The corresponding y-coordinate is $y_Q = -tx_Q = -t\left(\frac{4a}{t^2}\right) = -\frac{4a}{t}$. So, the coordinates of Q are $\left(\frac{4a}{t^2}, -\frac{4a}{t}\right)$.

The distance OQ from the origin O(0,0) to Q is: $OQ = \sqrt{\left(\frac{4a}{t^2}\right)^2 + \left(-\frac{4a}{t}\right)^2} = \sqrt{\frac{16a^2}{t^4} + \frac{16a^2}{t^2}} = \sqrt{\frac{16a^2(1+t^2)}{t^4}} = \frac{4|a|\sqrt{1+t^2}}{t^2}$

Now, we compute the product of OP and OQ: $OP \cdot OQ = \left(\frac{|a|t^2}{\sqrt{1+t^2}}\right) \cdot \left(\frac{4|a|\sqrt{1+t^2}}{t^2}\right) = 4|a|^2 = 4a^2$ This product is independent of the parameter $t$, hence it is a constant.