Explanation:
C Let the two tangent to the parabola \(y^2=4 \mathrm{ax}\) be PT and QT which are at right angle to one another at \(\mathrm{T}(\mathrm{e}, \mathrm{f})\). Then we have to find the locus of \(\mathrm{T}(\mathrm{e}, \mathrm{f})\).
We know that \(y=m x+\frac{a}{m}\), where \(m\) is the slope is the equation of tangent to the parabola \(\mathrm{y}^2=4 \mathrm{ax}\)
Since, this tangent to the parabola will pass through \(\mathrm{T}(\mathrm{e}, \mathrm{f})\) so
\(\mathrm{f}=\mathrm{me}+\frac{\mathrm{a}}{\mathrm{m}} ; \text { or } \mathrm{m}^2 \mathrm{e}-\mathrm{mf}+\mathrm{a}=0\)
This is a quadratic equation in \(\mathrm{m}\). So we have two roots, say \(\mathrm{m}_1\) and \(\mathrm{m}_2\), then
\(\mathrm{m}_1+\mathrm{m}_2=\frac{\mathrm{f}}{\mathrm{e}} \text {, and } \mathrm{m}_1 \times \mathrm{m}_2=\frac{\mathrm{a}}{\mathrm{e}}\)
\(\because\) The two tangents intersect at right angle so
\(\mathrm{m}_1 \cdot \mathrm{m}_2=-1 \text { or } \frac{\mathrm{a}}{\mathrm{e}}=-1 \text { or } \mathrm{h}+\mathrm{a}=0\)The locus of \(T(e, f)\) is \(x+a=0\), which is the equation of directrix.