Question

If \[{{r}_{1}}\] and \[{{r}_{2}}\] be the lengths of radii vectors of the parabola which are drawn at right angles to one another from the vertex, Prove that
\[{{r}_{1}}^{\dfrac{4}{3}}{{r}_{2}}^{\dfrac{4}{3}}=16{{a}^{2}}\left( {{r}_{1}}^{\dfrac{2}{3}}+{{r}_{2}}^{\dfrac{2}{3}} \right)\]

Answer 1

Hint: The coordinates of points where radius vector touches parabola, say $P$ and $Q$, can be obtained as \[\left( {{r}_{1}}\cos \theta ,{{r}_{1}}\sin \theta \right)\] and \[\left( {{r}_{2}}\sin \theta ,-{{r}_{2}}\cos \theta \right)\]. Then, these can be substituted in the equation of parabola to form equations.

Complete step by step answer:
Let us consider a parabola of the form \[{{y}^{2}}=4ax\].
Since \[{{r}_{1}}\] is a radius vector, let us consider that it makes an angle $\theta $ with the positive direction of \[x-\]axis and touches the parabola at point $P$.
So we can obtain the coordinates of point $P$ as \[\left( {{r}_{1}}\cos \theta ,{{r}_{1}}\sin \theta \right)\].
Now, it is said in the question that \[{{r}_{1}}\] and \[{{r}_{2}}\] are drawn at right angles to one another from the vertex. So, we can conclude that \[{{r}_{2}}\] makes an angle \[90-\theta \] with the positive direction of the \[x-\]axis and touches the parabola at point $Q$.
So we can obtain the coordinates of point $Q$ as \[\left( {{r}_{2}}\sin \theta ,-{{r}_{2}}\cos \theta \right)\].
Consider the figure as shown below.

Now, since $P$ and $Q$ lie on the parabola, they satisfy the equation of the parabola.
Therefore, we can substitute $P$\[\left( {{r}_{1}}\cos \theta ,{{r}_{1}}\sin \theta \right)\] in the parabola \[{{y}^{2}}=4ax\] as shown below,
\[\begin{align}
  & {{\left( {{r}_{1}}\sin \theta \right)}^{2}}=4a{{r}_{1}}\cos \theta \\
 & {{r}_{1}}^{2}{{\sin }^{2}}\theta =4a{{r}_{1}}\cos \theta \\
 & {{r}_{1}}{{\sin }^{2}}\theta =4a\cos \theta \ldots \ldots \ldots \left( i \right) \\
\end{align}\]
Now let us substitute the point $Q$\[\left( {{r}_{2}}\sin \theta ,-{{r}_{2}}\cos \theta \right)\]in the parabola \[{{y}^{2}}=4ax\] as shown below,
\[\begin{align}
  & {{\left( -{{r}_{2}}\cos \theta \right)}^{2}}=4a{{r}_{2}}\sin \theta \\
 & {{r}_{2}}^{2}{{\cos }^{2}}\theta =4a{{r}_{2}}\sin \theta \\
 & {{r}_{2}}{{\cos }^{2}}\theta =4a\sin \theta \ldots \ldots \ldots \left( ii \right) \\
\end{align}\]
Let us divide equation \[\left( i \right)\] by \[\left( ii \right)\]
$\dfrac{{{r}_{1}}{{\sin }^{2}}\theta }{{{r}_{2}}{{\cos }^{2}}\theta }=\dfrac{4a\cos \theta }{4a\sin \theta }$
On cancelling the like terms, we get
$\dfrac{{{r}_{1}}{{\sin }^{2}}\theta }{{{r}_{2}}{{\cos }^{2}}\theta }=\dfrac{\cos \theta }{\sin \theta }$
On rearranging the terms, we get
$\dfrac{{{\sin }^{3}}\theta }{{{\cos }^{3}}\theta }=\dfrac{{{r}_{2}}}{{{r}_{1}}}$
Since we know that $\dfrac{\sin \theta }{\cos \theta }=\tan \theta $, we can write
$\begin{align}
  & {{\tan }^{3}}\theta =\dfrac{{{r}_{2}}}{{{r}_{1}}} \\
 & \tan \theta ={{\left( \dfrac{{{r}_{2}}}{{{r}_{1}}} \right)}^{\dfrac{1}{3}}} \\
\end{align}$
We can use the trigonometric relation \[\cos \theta =\dfrac{1}{\sqrt{1+{{\tan }^{2}}\theta }}\] and substitute the above result in it.
\[\begin{align}
  & \cos \theta =\dfrac{1}{\sqrt{1+{{\left[ {{\left( \dfrac{{{r}_{2}}}{{{r}_{1}}} \right)}^{\dfrac{1}{3}}} \right]}^{2}}}} \\
 & \cos \theta =\dfrac{1}{\sqrt{1+{{\left( \dfrac{{{r}_{2}}}{{{r}_{1}}} \right)}^{\dfrac{2}{3}}}}} \\
\end{align}\]
Taking the LCM, we get
\[\begin{align}
  & \cos \theta =\dfrac{1}{\sqrt{\dfrac{{{r}_{1}}^{\dfrac{2}{3}}+{{r}_{2}}^{\dfrac{2}{3}}}{{{r}_{1}}^{\dfrac{2}{3}}}}} \\
 & \cos \theta =\dfrac{\sqrt{{{r}_{1}}^{\dfrac{2}{3}}}}{\sqrt{{{r}_{1}}^{\dfrac{2}{3}}+{{r}_{2}}^{\dfrac{2}{3}}}} \\
 & \cos \theta =\dfrac{{{r}_{1}}^{\dfrac{1}{3}}}{\sqrt{{{r}_{1}}^{\dfrac{2}{3}}+{{r}_{2}}^{\dfrac{2}{3}}}} \\
\end{align}\]
We can also use the trigonometric relation \[\sin \theta =\tan \theta \times \cos \theta \] and substitute $\tan \theta ={{\left( \dfrac{{{r}_{2}}}{{{r}_{1}}} \right)}^{\dfrac{1}{3}}}$ and \[\cos \theta =\dfrac{{{r}_{1}}^{\dfrac{1}{3}}}{\sqrt{{{r}_{1}}^{\dfrac{2}{3}}+{{r}_{2}}^{\dfrac{2}{3}}}}\] in it. So, we will get
$\begin{align}
  & \sin \theta ={{\left( \dfrac{{{r}_{2}}}{{{r}_{1}}} \right)}^{\dfrac{1}{3}}}\times \dfrac{{{r}_{1}}^{\dfrac{1}{3}}}{\sqrt{{{r}_{1}}^{\dfrac{2}{3}}+{{r}_{2}}^{\dfrac{2}{3}}}} \\
 & \sin \theta =\dfrac{{{r}_{2}}^{\dfrac{1}{3}}}{{{r}_{1}}^{\dfrac{1}{3}}}\times \dfrac{{{r}_{1}}^{\dfrac{1}{3}}}{\sqrt{{{r}_{1}}^{\dfrac{2}{3}}+{{r}_{2}}^{\dfrac{2}{3}}}} \\
 & \sin \theta =\dfrac{{{r}_{2}}^{\dfrac{1}{3}}}{\sqrt{{{r}_{1}}^{\dfrac{2}{3}}+{{r}_{2}}^{\dfrac{2}{3}}}} \\
\end{align}$
On substituting \[\cos \theta =\dfrac{{{r}_{1}}^{\dfrac{1}{3}}}{\sqrt{{{r}_{1}}^{\dfrac{2}{3}}+{{r}_{2}}^{\dfrac{2}{3}}}}\] and \[\sin \theta =\dfrac{{{r}_{2}}^{\dfrac{1}{3}}}{\sqrt{{{r}_{1}}^{\dfrac{2}{3}}+{{r}_{2}}^{\dfrac{2}{3}}}}\] in equation \[\left( i \right)\], we get
\[\begin{align}
  & {{r}_{1}}{{\left( \dfrac{{{r}_{2}}^{\dfrac{1}{3}}}{\sqrt{{{r}_{1}}^{\dfrac{2}{3}}+{{r}_{2}}^{\dfrac{2}{3}}}} \right)}^{2}}=4a\dfrac{{{r}_{1}}^{\dfrac{1}{3}}}{\sqrt{{{r}_{1}}^{\dfrac{2}{3}}+{{r}_{2}}^{\dfrac{2}{3}}}} \\
 & {{r}_{1}}\dfrac{{{r}_{2}}^{\dfrac{2}{3}}}{{{r}_{1}}^{\dfrac{2}{3}}+{{r}_{2}}^{\dfrac{2}{3}}}=4a\dfrac{{{r}_{1}}^{\dfrac{1}{3}}}{\sqrt{{{r}_{1}}^{\dfrac{2}{3}}+{{r}_{2}}^{\dfrac{2}{3}}}} \\
\end{align}\]
On rearranging the terms, we get
\[\begin{align}
  & {{r}_{1}}\dfrac{{{r}_{2}}^{\dfrac{2}{3}}}{{{r}_{1}}^{\dfrac{1}{3}}}=4a\dfrac{{{r}_{1}}^{\dfrac{2}{3}}+{{r}_{2}}^{\dfrac{2}{3}}}{\sqrt{{{r}_{1}}^{\dfrac{2}{3}}+{{r}_{2}}^{\dfrac{2}{3}}}} \\
 & {{r}_{1}}^{\left( 1-\dfrac{1}{3} \right)}{{r}_{2}}^{\dfrac{2}{3}}=4a\sqrt{{{r}_{1}}^{\dfrac{2}{3}}+{{r}_{2}}^{\dfrac{2}{3}}} \\
 & {{r}_{1}}^{\dfrac{2}{3}}{{r}_{2}}^{\dfrac{2}{3}}=4a\sqrt{{{r}_{1}}^{\dfrac{2}{3}}+{{r}_{2}}^{\dfrac{2}{3}}} \\
\end{align}\]
Squaring both sides, we get
\[\begin{align}
  & {{\left( {{r}_{1}}^{\dfrac{2}{3}}{{r}_{2}}^{\dfrac{2}{3}} \right)}^{2}}={{\left[ 4a\sqrt{{{r}_{1}}^{\dfrac{2}{3}}+{{r}_{2}}^{\dfrac{2}{3}}} \right]}^{2}} \\
 & {{r}_{1}}^{\dfrac{4}{3}}{{r}_{2}}^{\dfrac{4}{3}}=16{{a}^{2}}\left( {{r}_{1}}^{\dfrac{2}{3}}+{{r}_{2}}^{\dfrac{2}{3}} \right) \\
\end{align}\]

Therefore, we have proved that if \[{{r}_{1}}\] and \[{{r}_{2}}\] be the lengths of radii vectors of the parabola which are drawn at right angles to one another from the vertex, \[{{r}_{1}}^{\dfrac{4}{3}}{{r}_{2}}^{\dfrac{4}{3}}=16{{a}^{2}}\left( {{r}_{1}}^{\dfrac{2}{3}}+{{r}_{2}}^{\dfrac{2}{3}} \right)\].

Note: The choice of the parabola i.e. either \[{{y}^{2}}=4ax\] or \[{{x}^{2}}=4ay\] is important. Great attention must be paid as the points chosen will change accordingly. For the parabola, \[{{x}^{2}}=4ay\], the coordinates of points $P$ and $Q$ would be \[\left( {{r}_{1}}\cos \theta ,{{r}_{1}}\sin \theta \right)\] and \[\left( -{{r}_{2}}\cos \theta ,{{r}_{2}}\sin \theta \right)\] respectively.