The locus of the point of intersection of two tangents to the...

Let the two tangents to the parabola $y^{2}=4 a x$ be $P T$ and $Q T$ which are at right angle to one another at ($\mathrm{T}(\mathrm{h}, \mathrm{k})$). Then we have a find the locus of ($\mathrm{T}(\mathrm{h}, \mathrm{k})$). We know that (y=m x+$\frac{a}{m}$), where $m$ is the slope is the equation of tangent to the parabola (\mathrmy^{2}=4 $\mathrm{ax}$) for all ($\mathrm{m}$). Since this tangent to the parabola will pass through ($\mathrm{T}(\mathrm{h}, \mathrm{k})$), so ($\mathrm{k}=\mathrm{mh}+\frac{\mathrm{a}}{\mathrm{m}}$; $\text{ or } \mathrmm^{2} \mathrm{~h}-\mathrm{mk}+\mathrm{a}=$0) This is a quadratic equation in $m$, so will have two roots, say $m_{1}$ and $m_{2}$, then ($\mathrm{m}_{1}+\mathrm{m}_{2}=\frac{\mathrm{k}}{\mathrm{h}}$), and ($\mathrm{m}_{1} \cdot \mathrm{m}_{2}=\frac{\mathrm{a}}{\mathrm{h}}$) Given that the two tangents intersect at right angle so ($\mathrm{m}_{1} \cdot \mathrm{m}_{2}=-$1 $\text{ or } \frac{\mathrm{a}}{\mathrm{h}}=-$1 $\text{ or } \mathrm{h}+\mathrm{a}=$0) The locus of ($\mathrm{T}(\mathrm{h}, \mathrm{k})$) is ($\mathrm{x}+\mathrm{a}=$0), which is the equation of directrix.