Question

PT is the tangent at any point P on the parabola ${{y}^{2}}=4ax$ and SP is the focal distance and PM is the perpendicular from P to the directrix, then
(a) \[\angle \] MPT = 2 \[\angle \] SPT
(b) \[\angle \] MPT = \[\angle \] SPT
(c) \[\angle \] MPT = $\dfrac{1}{2}$ \[\angle \] SPT
(d) None of these

Answer 1

Hint: To solve the above problem, we will use the concept of the parametric equation of parabola which is of the form $x=a{{t}^{2}},y=2at$, and then taking point P as the parametric point of the parabola we get the equation of tangent PT at that point. We draw a perpendicular PM from point P to the directrix. As we know that, suppose A$\left( {{x}_{1}},{{y}_{1}} \right)$ and B $\left( {{x}_{2}},{{y}_{2}} \right)$ are two points then slope of AB is given by $\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$.So we will find the slopes of PM, PT, and SP where S is the focus of a parabola, using the above formula and then using the formula to find the angle between two lines that is, if the slope of two lines is supposed m and ${{m}_{0}}$ then the angle between these two lines are given by $\tan \theta =\dfrac{(m-{{m}_{0}})}{(1+m{{m}_{0}})}$ where $\theta $ is the angle between the lines., using this formula we find the angles between PM and PT as \[\angle \] MPT and between SP and PT as \[\angle \] SPT then we conclude the relationship between these two angles.

Complete step by step answer:
The equation of a given parabola is ${{y}^{2}}=4ax$
Now let us assume that P $(a{{t}^{2}},2\text{at})$ be a parametric point on the given parabola ${{y}^{2}}=4ax$. Generally, equation of tangent to a parabola ${{y}^{2}}=4ax$ at $({{x}_{1}},{{y}_{1}})$ is given by $y{{y}_{1}}-2a(x+{{x}_{1}})=0$ , so here ${{x}_{1}}=a{{t}^{2}}\text{ and }{{y}_{1}}=2at$ , so the equation of tangent to the parabola at point P is given by $y(2at)-2(a)(x+\text{a}{{\text{t}}^{2}})=0$ i.e. $x-yt+a{{t}^{2}}=0\text{ or }y=\dfrac{x}{t}+at$ .
 Now, we have slope of PT from given equation, (say) ${{m}_{1}}=\dfrac{1}{t}$ , slope of PM (say)${{m}_{2}}=\dfrac{2at-2at}{a{{t}^{2}}-a}=0$ and slope of SP (say) ${{m}_{3}}=\dfrac{2at-0}{a{{t}^{2}}-a}=\dfrac{2at}{a({{t}^{2}}-1)}=\dfrac{2t}{{{t}^{2}}-1}$ .
Now, using the formula to find the angle between two lines that is, if the slope of two lines is supposed m and ${{m}_{0}}$ then the angle between these two lines are given by $\tan \theta =\dfrac{(m-{{m}_{0}})}{(1+m{{m}_{0}})}$ where $\theta $ is the angle between the lines.
So, let us assume here the angle between the PM and PT be ${{\theta }_{1}}\text{ i}\text{.e}\text{. }\angle \text{ MPT}={{\theta }_{1}}$ and angle between PT and SP be ${{\theta }_{2}}\text{ i}\text{.e}\text{. }\angle \,\text{SPT}={{\theta }_{2}}$ , then we have $\tan {{\theta }_{1}}=\dfrac{({{m}_{1}}-{{m}_{2}})}{(1+{{m}_{1}}{{m}_{2}})}=\dfrac{\left( \dfrac{1}{t}-0 \right)}{\left( 1+\dfrac{1}{t}(0) \right)}=\dfrac{1}{t}\Rightarrow {{\theta }_{1}}={{\tan }^{-1}}\dfrac{1}{t}$ …..(i) and$\begin{align}
  & \tan {{\theta }_{2}}=\dfrac{\left( {{m}_{3}}-{{m}_{1}} \right)}{\left( 1+{{m}_{1}}{{m}_{3}} \right)}=\dfrac{\left( \dfrac{2t}{{{t}^{2}}-1}-\dfrac{1}{t} \right)}{\left( 1+\dfrac{2t}{{{t}^{2}}-1}\cdot \dfrac{1}{t} \right)}=\dfrac{\left( \dfrac{2{{t}^{2}}-{{t}^{2}}+1}{t({{t}^{2}}-1)} \right)}{\left( \dfrac{t({{t}^{2}}-1)+2t}{t({{t}^{2}}-1)} \right)}=\dfrac{{{t}^{2}}+1}{t({{t}^{2}}-1+2)}=\dfrac{{{t}^{2}}+1}{t({{t}^{2}}+1)}=\dfrac{1}{t} \\
 & \Rightarrow {{\theta }_{2}}={{\tan }^{-1}}\dfrac{1}{t}\,\,\,...........(ii) \\
\end{align}$
Now, from equation (i) and (ii), we get that
${{\theta }_{1}}={{\theta }_{2}}\Rightarrow \,\,\angle \,\,\text{MPT}=\angle \,\,\text{SPT}$ .
Therefore, the correct answer is an option (b)

Note:
In order to solve any conic section problem, always try to draw a proper diagram, required as per the question, and then proceed further, and must take care of the points to be drawn in the diagram. And always try to attack the question with the parametric form if the point at which tangent is to be drawn is not given. And also while calculating slopes for any two points suppose A$\left( {{x}_{1}},{{y}_{1}} \right)$ and B $\left( {{x}_{2}},{{y}_{2}} \right)$,don’t confuse to write slope of AB as \[\dfrac{{{x}_{2}}-{{x}_{1}}}{{{y}_{2}}-{{y}_{1}}}\], this is the common mistake one does, where the correct formula to find slope of AB is $\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$