Question

A normal chord of a parabola subtending a right angle at the vertex makes an acute angle \[\theta \] with the X – axis, then the value of \[\theta \] is
(a) \[\arctan 2\]
(b) \[\arctan \sqrt{2}\]
(c) \[\text{arccot} \sqrt{2}\]
(d) None of the above

Answer 1

Hint: We solve this problem by assuming the equation of given parabola as standard equation of parabola that is \[{{y}^{2}}=4ax\]
We use the condition that the equation of normal at some parametric point \[\left( a{{t}^{2}},2at \right)\] of the parabola \[{{y}^{2}}=4ax\]is given as
\[y=-tx+2at+a{{t}^{3}}\]
Then we use the condition two lines of slopes \[{{m}_{1}},{{m}_{2}}\] forms a right angle then
\[{{m}_{1}}\times {{m}_{2}}=-1\]
We also have other condition that if a line makes an angle \[\alpha \] with the X – axis in anti – clockwise direction then the slope of line is given as
\[m=\tan \alpha \]
By using the above two formulas to given conditions we find the required value.


Complete step by step answer:
We are given that there is a normal chord that subtends a right angle at the vertex.
Let us assume that the equation of given parabola as standard equation of parabola that is \[{{y}^{2}}=4ax\]
Now, let us assume that the normal is drawn at a parametric point \[P\left( a{{t}_{1}}^{2},2a{{t}_{1}} \right)\] such that it touches the parabola again at the point \[Q\left( a{{t}_{2}}^{2},2a{{t}_{2}} \right)\]
Now, let us take a rough figure of the given information as follows

We know that the condition that the equation of normal at some parametric point \[\left( a{{t}^{2}},2at \right)\] of the parabola \[{{y}^{2}}=4ax\]is given as
\[y=-tx+2at+a{{t}^{3}}\]
By using the above formula to normal PQ from point P then we get the equation of PQ as
\[y=-{{t}_{1}}x+2a{{t}_{1}}+a{{t}_{1}}^{3}\]
Here, we can see that the slope of the line PQ is given as
\[\Rightarrow {{m}_{PQ}}={{t}_{1}}\]
We are given that the normal makes an acute angle with X –axis as shown in the figure.
We know that if a line makes an angle \[\alpha \] with the X – axis in anti – clockwise direction then the slope of line is given as
\[m=\tan \alpha \]
By using the above formula to given angle then we get
\[\begin{align}
  & \Rightarrow {{m}_{PQ}}=\tan \left( {{180}^{\circ }}-\theta \right) \\
 & \Rightarrow \tan \theta =-{{t}_{1}}............equation(i) \\
\end{align}\]
Here, we can see that Q lies in the equation of PQ.
So, by substituting the point Q in equation of PQ then we get
\[\begin{align}
  & \Rightarrow 2a{{t}_{2}}=-{{t}_{1}}\left( a{{t}_{2}}^{2} \right)+2a{{t}_{1}}+a{{t}_{1}}^{3} \\
 & \Rightarrow 2{{t}_{2}}-2{{t}_{1}}=-{{t}_{1}}\left( {{t}_{2}}^{2}-{{t}_{1}}^{2} \right) \\
\end{align}\]
By using the formula of algebra that is
\[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\]
To above equation then we get
\[\begin{align}
  & \Rightarrow 2\left( {{t}_{2}}-{{t}_{1}} \right)=-{{t}_{1}}\left( {{t}_{2}}+{{t}_{1}} \right)\left( {{t}_{2}}-{{t}_{1}} \right) \\
 & \Rightarrow 2=-{{t}_{1}}{{t}_{2}}-{{t}_{1}}^{2}.............equation(ii) \\
\end{align}\]
Now, let us use the condition that PQ makes right angle at vertex
We know that the formula of slope of two points \[\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\] is given as
\[m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]
We also know that the condition two lines of slopes \[{{m}_{1}},{{m}_{2}}\] forms a right angle then
\[{{m}_{1}}\times {{m}_{2}}=-1\]
By using the above two formulas to OP and OQ then we get
\[\begin{align}
  & \Rightarrow {{m}_{OP}}\times {{m}_{OQ}}=-1 \\
 & \Rightarrow \left( \dfrac{2a{{t}_{1}}-0}{a{{t}_{1}}^{2}-0} \right)\left( \dfrac{2a{{t}_{2}}-0}{a{{t}_{2}}^{2}-0} \right)=-1 \\
 & \Rightarrow \dfrac{4}{{{t}_{1}}{{t}_{2}}}=-1 \\
 & \Rightarrow {{t}_{1}}{{t}_{2}}=-4 \\
\end{align}\]
By substituting the above result in equation (ii) then we get
\[\begin{align}
  & \Rightarrow 2=-4-{{t}_{1}}^{2} \\
 & \Rightarrow {{t}_{1}}^{2}=2 \\
 & \Rightarrow {{t}_{1}}=\pm \sqrt{2} \\
\end{align}\]
By substituting the above result in equation (i) then we get
\[\Rightarrow \tan \theta =-\left( \pm \sqrt{2} \right)\]
We know that \[\theta \] is acute angle so that tangent should be positive.
By using the above condition we get
\[\Rightarrow \theta =\arctan \sqrt{2}\]
Therefore, we can conclude that the value of \[\theta \] is \[\arctan \sqrt{2}\]
So, option (a) is correct answer.

Note:
 Students may do mistakes in taking the \[\theta \] given with the slope of the line.
We have the condition that if a line makes an angle \[\alpha \] with the X-axis in an anti-clockwise direction then the slope of the line is given as
\[m=\tan \alpha \]
By using the above condition to the above figure then we get the equation as
\[\Rightarrow {{m}_{PQ}}=\tan \left( {{180}^{\circ }}-\theta \right)\]
But students may miss the direction of the angle to be taken and assume that angle as
\[\Rightarrow {{m}_{PQ}}=\tan \left( \theta \right)\]