The locus of the point of intersection of two tangents to the parabola $y^{2}=4 a x,$ which are at right…
The locus of the point of intersection of two tangents to the parabola $y^{2}=4 a x,$ which are at right angle to one another is
$x^{2}+y^{2}=a^{2}$
$a y^{2}=x$
$x+a=0$
$x+y \pm a=0$
Solution
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 \(\mathrm{y}^{2}=4 \mathrm{ax}\) for all \(\mathrm{m}\).
Since this tangent to the parabola will pass through \(\mathrm{T}(\mathrm{h}, \mathrm{k})\), so
\(\mathrm{k}=\mathrm{mh}+\frac{\mathrm{a}}{\mathrm{m}} ; \text { or } \mathrm{m}^{2} \mathrm{~h}-\mathrm{mk}+\mathrm{a}=0\)
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.
