Question

If a normal chord subtends a right angle at the vertex of the parabola \[{{y}^{2}}=4ax\], prove that it is inclined at an angle of \[{{\tan }^{-1}}\left( \sqrt{2} \right)\] to the axis of the parabola.

Answer 1

Hint: Suppose two points of the normal chord lying on parabola as \[\left( at_{1}^{2},2a{{t}_{1}} \right)\] and \[\left( at_{2}^{2},2a{{t}_{2}} \right)\]. Let the slope of tangent at the point where the chord acts as a normal for parabola, by differentiating the curve, \[{{y}^{2}}=4ax\] at that point. Hence, get the slope of normal using relation.

Complete step-by-step answer:
Product of slopes of two perpendicular lines = -1.
Get the slope of chord using the coordinates supposed as well with the help of relation \[=\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\], where \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] are lying on the line. And use the given condition to solve the problem further.

As, we need to prove the angle formed (inclination) with the axis of parabola, \[{{y}^{2}}=4ax\] to \[{{\tan }^{-1}}\left( \sqrt{2} \right)\] with the help of given in formations in the problem.
We know the axis of parabola, \[{{y}^{2}}=4ax\] is x – axis, as it is symmetric about x – axis. So, we need to determine the angle with the x – axis and prove it to \[{{\tan }^{-1}}\left( \sqrt{2} \right)\]. So, we can calculate the slope of the chord and equate it to the \[\tan \theta \] because slope is defined as tan of angle formed with the positive direction of the x – axis.
So, let us suppose slope of normal chord \[=\tan \theta \]
As, we need to find the slope of a chord in \[{{y}^{2}}=4ax\], which is normal at one end and subtends a right angle at the origin as well. So, let the two ends of the chord are \[\left( at_{1}^{2},2a{{t}_{1}} \right)\] and \[\left( at_{2}^{2},2a{{t}_{2}} \right)\]. [Parametric coordinates for \[{{y}^{2}}=4ax\]].
So, diagram can be represented as,

   

Let AB is acting as a normal at the point B.
We know the slope of tangent at point B can be given as,
\[{{\left. \dfrac{dy}{dx} \right|}_{\left( at_{2}^{2},2a{{t}_{2}} \right)}}-(i)\]
Where, we need to use relation, \[{{y}^{2}}=4ax\].
So, differentiating \[{{y}^{2}}=4ax\], we get,
\[\begin{align}
  & \dfrac{d}{dx}{{y}^{2}}=\dfrac{d}{dx}\left( 4a.x \right) \\
 & 2y\dfrac{dy}{dx}=4a\times 1 \\
\end{align}\]
Where, we know,
\[\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}\]
Hence, we get,
\[\begin{align}
  & y\dfrac{dy}{dx}=2a \\
 & \dfrac{dy}{dx}=\dfrac{2a}{y} \\
\end{align}\]
Now, we can get slope of tangent at point B as,
\[{{\left. \dfrac{dy}{dx} \right|}_{\left( at_{2}^{2},2a{{t}_{2}} \right)}}=\dfrac{2a}{2a{{t}_{2}}}=\dfrac{1}{{{t}_{2}}}-(ii)\]
Now, we know tangent is perpendicular to the normal at the point of tangency for any conic. And as, we know the relation between slope of two perpendicular lines is given as,
Product of two perpendicular lines = -1 – (iii)
So, we get,
Slope of normal at BX slope of tangent at B = -1.
\[\dfrac{1}{{{t}_{2}}}\times \] slope of normal at B = -1
Slope of normal at B = \[-{{t}_{2}}\]
Now, we can calculate slope of normal at B i.e. slope of line AB by formula,
Slope = \[\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}-(iv)\]
So, we get,
Slope of normal at B \[=\dfrac{2a{{t}_{2}}-2a{{t}_{1}}}{at_{2}^{2}-at_{1}^{2}}\]
\[\begin{align}
  & =\dfrac{2a\left( {{t}_{2}}-{{t}_{1}} \right)}{a\left( t_{2}^{2}-t_{1}^{2} \right)} \\
 & =\dfrac{2\left( {{t}_{2}}-{{t}_{1}} \right)}{\left( {{t}_{2}}+{{t}_{1}} \right)\left( {{t}_{2}}-{{t}_{1}} \right)} \\
\end{align}\]
Slope of normal at B = \[\dfrac{2}{\left( {{t}_{2}}+{{t}_{1}} \right)}\]
As, we have already calculated the slope of normal at point B as \[-{{t}_{2}}\].
 So, we get, \[\dfrac{2}{{{t}_{1}}+{{t}_{2}}}=-{{t}_{2}}\]
\[\begin{align}
  & -2={{t}_{2}}\left( {{t}_{1}}+{{t}_{2}} \right) \\
 & {{t}_{1}}{{t}_{2}}+t_{2}^{2}=-2-(v) \\
\end{align}\]
Now, as AO and BO are perpendicular to each other. So, we can calculate the slopes of them and use equation (iii).
So, we get slope of AO from the equation (iv) as,
Slope of AO = \[\dfrac{2a{{t}_{1}}-0}{at_{1}^{2}-0}=\dfrac{2a{{t}_{1}}}{at_{1}^{2}}\]
Slope of AO \[=\dfrac{2}{{{t}_{1}}}\]
Similarly, slope of BO \[=\dfrac{2}{{{t}_{2}}}\]
Now, using equation (iii), we get,
\[\begin{align}
  & \dfrac{2}{{{t}_{1}}}\times \dfrac{2}{{{t}_{2}}}=-1 \\
 & 4=-{{t}_{1}}{{t}_{2}} \\
\end{align}\]
Or
\[{{t}_{1}}{{t}_{2}}=-4-(vi)\]
Now, putting, \[{{t}_{1}}{{t}_{2}}=-4\] to the equation (v), we get,
\[\begin{align}
  & -4+t_{2}^{2}=-2 \\
 & t_{2}^{2}=4-2=2 \\
 & {{t}_{2}}=\pm \sqrt{2} \\
\end{align}\]
We know the slope of the chord is given as \[-{{t}_{2}}\]. So, the possible slope of the chord is \[\pm \sqrt{2}\].
Hence, \[\pm \sqrt{2}\] is a slope of the normal chord as well.
The slope of the chord is \[\tan \theta \]. So, we can equate \[\pm \sqrt{2}\] to \[\tan \theta \].
So, we get,
\[\begin{align}
  & \tan \theta =\pm \sqrt{2} \\
 & \theta ={{\tan }^{-1}}\left( \pm \sqrt{2} \right) \\
\end{align}\]
As, there are two inclinations of the chord with the x – axis (one with positive direction of x – axis and another with negative direction).
So, we will get two angles by the same chord by the equation, \[\theta ={{\tan }^{-1}}\left( \pm \sqrt{2} \right)\].
Hence, the inclination can also be said as \[{{\tan }^{-1}}\left( \sqrt{2} \right)\] only as well. So, the statement given in the problem is proved.

Note: One may use the direct result that is two points \[\left( at_{1}^{2},2a{{t}_{1}} \right)\] and \[\left( at_{2}^{2},2a{{t}_{2}} \right)\] is subtending \[{{90}^{\circ }}\] at vertex of parabola then, \[{{t}_{1}}{{t}_{2}}=-4\]. So, one may use this relation for future reference as well.
One may use points \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] as well for representing the points A and B. And to need to use two more equations, that are \[y_{1}^{2}=4a{{x}_{1}}\] and \[y_{2}^{2}=4a{{x}_{2}}\]. So, it can be another approach. But the involvement of two more equations, may make the problem complex for some of the students. So, always try to use parametric form of coordinates on conics with these kinds of problems.
We can use another approach to get the slope of the normal at any conic as well. We can write the equation of tangent at point A and B by using equation, T = 0.
i.e. replace \[{{x}^{2}}\] by \[x{{x}_{1}}\], \[{{y}^{2}}\] by \[y{{y}_{1}}\].
x by \[\dfrac{x+{{x}_{1}}}{2}\], y by \[\dfrac{y+{{y}_{1}}}{2}\], where \[\left( {{x}_{1}},{{y}_{1}} \right)\] is the point on the given sonic. Hence, compare the calculated equation of tangent with the line \[y=mx+c\] to get slope of tangent and hence, get slope of normal by relation.
Product of slopes of two perpendicular lines = -1.