Question

If $\omega$ be the angle which a focal chord of a parabola makes with the axis, prove that the length of the chord is $4a{{\operatorname{cosec}}^{2}}\omega$ and that the perpendicular on it from the vertex is $a\sin \omega$.

Answer 1

We start solving the problem by assuming the equation of parabola and ends of the focal chord. We then find the equation of the focal chord using these points and compare with the standard form$y=mx+c$. We then use the fact that the slope of the line is defined as tangent of the angle made by the line with x-axis to get the difference relations between the coordinates of the assumed points. We then use this relationship to find the length of the focal chord and perpendicular distance from vertex to the focal chord.

Complete step by step answer:
According to the problem, we are given that $\omega$ be the angle which a focal chord of a parabola makes with the axis and we need to prove that the length of the chord is $4a{{\operatorname{cosec}}^{2}}\omega$ and that the perpendicular on it from the vertex is $a\sin \omega$.
Let us assume the equation of parabola be ${{y}^{2}}=4ax$ and ends of the focal chord in the parabola is $A\left( at_{1}^{2},2a{{t}_{1}} \right)$ and $B\left( at_{2}^{2},2a{{t}_{2}} \right)$. We know that for a focal chord ${{t}_{1}}{{t}_{2}}=-1$ ---(1).

Let us find the equation of the focal chord.
We know that the equation of the line passing through the point $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is $y-{{y}_{1}}=\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\times \left( x-{{x}_{1}} \right)$.
So, we get the equation of focal chord as $y-2a{{t}_{1}}=\left( \dfrac{2a{{t}_{2}}-2a{{t}_{1}}}{at_{2}^{2}-at_{1}^{2}} \right)\times \left( x-at_{1}^{2} \right)$.
Cancelling out similar terms and expanding the denominator, we have
$\Rightarrow y-2a{{t}_{1}}=\left( \dfrac{2\left( {{t}_{2}}-{{t}_{1}} \right)}{\left( {{t}_{2}}-{{t}_{1}} \right)\left( {{t}_{2}}+{{t}_{1}} \right)} \right)\times \left( x-at_{1}^{2} \right)$.
$\Rightarrow y-2a{{t}_{1}}=\left( \dfrac{2}{\left( {{t}_{2}}+{{t}_{1}} \right)} \right)\times \left( x-at_{1}^{2} \right)$.
$\Rightarrow y-2a{{t}_{1}}=\left( \dfrac{2}{\left( {{t}_{2}}+{{t}_{1}} \right)} \right)x-\left( \dfrac{2at_{1}^{2}}{\left( {{t}_{2}}+{{t}_{1}} \right)} \right)$.
$\Rightarrow y=\left( \dfrac{2}{\left( {{t}_{2}}+{{t}_{1}} \right)} \right)x-\left( \dfrac{2at_{1}^{2}}{\left( {{t}_{2}}+{{t}_{1}} \right)} \right)+2a{{t}_{1}}$.
Taking the LCM of the terms, we have
$\Rightarrow y=\left( \dfrac{2}{\left( {{t}_{2}}+{{t}_{1}} \right)} \right)x-\dfrac{2at_{1}^{2}}{\left( {{t}_{2}}+{{t}_{1}} \right)}+\dfrac{2at_{1}^{2}+2a{{t}_{1}}{{t}_{2}}}{\left( {{t}_{2}}+{{t}_{1}} \right)}$.
$\Rightarrow y=\left( \dfrac{2}{\left( {{t}_{2}}+{{t}_{1}} \right)} \right)x+\dfrac{-2at_{1}^{2}+2at_{1}^{2}+2a{{t}_{1}}{{t}_{2}}}{\left( {{t}_{2}}+{{t}_{1}} \right)}$.
$\Rightarrow y=\left( \dfrac{2}{\left( {{t}_{2}}+{{t}_{1}} \right)} \right)x+\left( \dfrac{2a{{t}_{1}}{{t}_{2}}}{\left( {{t}_{2}}+{{t}_{1}} \right)} \right)$.
From equation (1), we get equation of focal chord as $y=\left( \dfrac{2}{\left( {{t}_{2}}+{{t}_{1}} \right)} \right)x-\dfrac{2a}{\left( {{t}_{2}}+{{t}_{1}} \right)}$ ---(2).
Comparing with the standard equation $y=mx+c$, we get slope as $m=\dfrac{2}{{{t}_{1}}+{{t}_{2}}}$.
We know that the slope of the line is defined as the tangent of the angle made by the line with x-axis.
Since the focal chord makes an angle $\omega$ with the x-axis. So, we get $m=\tan \omega$.
So, we get $\tan \omega =\dfrac{2}{{{t}_{1}}+{{t}_{2}}}$ ---(3). Let us substitute equation (3) in equation (2).
So, we get the equation of the focal chord as $y=\tan \omega x-a\tan \omega$ ---(4).
From equation (3), we have ${{t}_{1}}+{{t}_{2}}=\dfrac{2}{\tan \omega }$.
$\Rightarrow {{t}_{1}}+{{t}_{2}}=2\cot \omega$ ---(5).
Now, let us square on both sides.
So, we get ${{\left( {{t}_{1}}+{{t}_{2}} \right)}^{2}}={{\left( 2\cot \omega \right)}^{2}}$.
$\Rightarrow {{\left( {{t}_{1}}+{{t}_{2}} \right)}^{2}}=4{{\cot }^{2}}\omega$.
$\Rightarrow {{\left( {{t}_{1}}-{{t}_{2}} \right)}^{2}}+4{{t}_{1}}{{t}_{2}}=4{{\cot }^{2}}\omega$.
From equation (1), we get ${{\left( {{t}_{1}}-{{t}_{2}} \right)}^{2}}-4=4{{\cot }^{2}}\omega$.
$\Rightarrow {{\left( {{t}_{1}}-{{t}_{2}} \right)}^{2}}=4+4{{\cot }^{2}}\omega$.
$\Rightarrow {{\left( {{t}_{1}}-{{t}_{2}} \right)}^{2}}=4\left( 1+{{\cot }^{2}}\omega \right)$.
$\Rightarrow {{\left( {{t}_{1}}-{{t}_{2}} \right)}^{2}}=4\left( {{\operatorname{cosec}}^{2}}\omega \right)$.
$\Rightarrow {{t}_{1}}-{{t}_{2}}=2\operatorname{cosec}\omega$ ---(6)
Let us find the length of the focal chord i.e., the distance between the points $A\left( at_{1}^{2},2a{{t}_{1}} \right)$ and $B\left( at_{2}^{2},2a{{t}_{2}} \right)$.
So, the length of focal chord = $\sqrt{{{\left( at_{2}^{2}-at_{1}^{2} \right)}^{2}}+{{\left( 2a{{t}_{2}}-2a{{t}_{1}} \right)}^{2}}}$.
Length of focal chord = $\sqrt{{{\left( a\left( {{t}_{2}}-{{t}_{1}} \right)\left( {{t}_{2}}+{{t}_{1}} \right) \right)}^{2}}+{{\left( 2a\left( {{t}_{2}}-{{t}_{1}} \right) \right)}^{2}}}$.
Length of focal chord = $\sqrt{{{a}^{2}}{{\left( {{t}_{2}}-{{t}_{1}} \right)}^{2}}\left( {{\left( {{t}_{2}}+{{t}_{1}} \right)}^{2}}+{{\left( 2 \right)}^{2}} \right)}$.
From equations (5) and (6), we get
Length of focal chord = $\sqrt{{{a}^{2}}{{\left( 2\operatorname{cosec}\omega \right)}^{2}}\left( {{\left( 2\cot \omega \right)}^{2}}+{{\left( 2 \right)}^{2}} \right)}$.
Length of focal chord = $\sqrt{4{{a}^{2}}{{\operatorname{cosec}}^{2}}\omega \left( 4{{\cot }^{2}}\omega +4 \right)}$.
Length of focal chord = $\sqrt{4{{a}^{2}}{{\operatorname{cosec}}^{2}}\omega \left( 4{{\operatorname{cosec}}^{2}}\omega \right)}$.
Length of focal chord = $\sqrt{16{{a}^{2}}{{\operatorname{cosec}}^{4}}\omega }$.
Length of focal chord = $4a{{\operatorname{cosec}}^{2}}\omega$.
So, we have proved that the length of the focal chord is $4a{{\operatorname{cosec}}^{2}}\omega$.
Now, we know that the vertex of the parabola ${{y}^{2}}=4ax$ is $\left( 0,0 \right)$. Let us find the perpendicular distance from $\left( 0,0 \right)$ to the focal chord $y=\tan \omega x-a\tan \omega \Leftrightarrow \tan \omega x-y-a\tan \omega =0$.
We know that the perpendicular distance from the $\left( p,q \right)$ to the line $ax+by+c=0$ is $\dfrac{\left| ap+bq+c \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$.
So, the perpendicular distance from $\left( 0,0 \right)$ to the focal chord $\tan \omega x-y-a\tan \omega =0$ is $\dfrac{\left| \tan \omega \left( 0 \right)-0-a\tan \omega \right|}{\sqrt{{{\tan }^{2}}\omega +{{\left( -1 \right)}^{2}}}}$.
$\Rightarrow \dfrac{\left| \tan \omega \left( 0 \right)-0-a\tan \omega \right|}{\sqrt{{{\tan }^{2}}\omega +{{\left( -1 \right)}^{2}}}}=\dfrac{\left| 0-a\tan \omega \right|}{\sqrt{{{\tan }^{2}}\omega +1}}$.
$\Rightarrow \dfrac{\left| \tan \omega \left( 0 \right)-0-a\tan \omega \right|}{\sqrt{{{\tan }^{2}}\omega +{{\left( -1 \right)}^{2}}}}=\dfrac{a\tan \omega }{\sqrt{{{\sec }^{2}}\omega }}$.
$\Rightarrow \dfrac{\left| \tan \omega \left( 0 \right)-0-a\tan \omega \right|}{\sqrt{{{\tan }^{2}}\omega +{{\left( -1 \right)}^{2}}}}=\dfrac{a\tan \omega }{\sec \omega }$.
$\Rightarrow \dfrac{\left| \tan \omega \left( 0 \right)-0-a\tan \omega \right|}{\sqrt{{{\tan }^{2}}\omega +{{\left( -1 \right)}^{2}}}}=\dfrac{\dfrac{a\sin \omega }{\cos \omega }}{\dfrac{1}{\cos \omega }}$.
$\Rightarrow \dfrac{\left| \tan \omega \left( 0 \right)-0-a\tan \omega \right|}{\sqrt{{{\tan }^{2}}\omega +{{\left( -1 \right)}^{2}}}}=a\sin \omega$.

So, we have found the perpendicular distance on the focal chord from the vertex as $a\sin \omega$.

Note: We can see that the problems contain heavy amounts of calculation this may give us confusion and will make us perform mistakes. We should know that the focal chord passes through the focus of the parabola and the latus rectum is one of the focal-chord. We can also assume the ends of the latus rectum to prove the required results of the problem.