Question

Find the locus of the middle points of chords of the parabola which subtend a constant angle \[\alpha \] at the vertex.

Answer 1

Hint: Since vertex is at origin the lines passing through it are of form \[y=mx\]. Take two general points on parabola and make lines passing through them and the origin. The angle between these two lines should be \[\alpha \].

Complete step-by-step answer:

Consider the above picture. \[A\left( a{{t}^{2}},2at \right)\]and\[B\left( a{{u}^{2}},2au \right)\]are two variable points on the standard parabola \[{{y}^{2}}=4ax\] with parameters \[t\] and \[u\] respectively.
We have been given that \[\angle AOB\] is constant i.e. equal to \[\alpha \].
We can also write \[\alpha \] as the angle between line \[AO\] and line \[BO\].
These are lines passing through origin and can be written as \[y=mx\] where \[m\] is the slope of the line.
So let,
\[AO\] be \[y=\dfrac{2}{t}x\]
\[BO\] be \[y=\dfrac{2}{u}x\]
(Slope of lines passing through origin are just ratio of \[y\] coordinate and \[x\] coordinate of any point on the line)
So,
$\tan \alpha =\dfrac{\dfrac{2}{t}-\dfrac{2}{u}}{1+\dfrac{2}{t}\times \dfrac{2}{u}}$
$\tan \alpha =\dfrac{\dfrac{2u}{ut}-\dfrac{2t}{ut}}{1+\dfrac{4}{ut}}$
$\tan \alpha =\dfrac{\dfrac{2u}{ut}-\dfrac{2t}{ut}}{\dfrac{ut}{ut}+\dfrac{4}{ut}}$
$\tan \alpha =\dfrac{\dfrac{2u-2t}{ut}}{\dfrac{ut+4}{ut}}$
$\tan \alpha =\dfrac{2\left( u-t \right)}{ut+4}...(i)$
Now let point \[\left( h,k \right)\]lie on our locus. Since they lie on the locus it is the midpoint of our variable chord AB.
So using the midpoint formula which is,
Xmid=\[\dfrac{x1+x2}{2}\]
ymid= \[\dfrac{y1+y2}{2}\]
We write,
$h=\left\{ \dfrac{\left( x~coordinate~of~A \right)+\left( x~coordinate~of~B \right)}{2} \right\}$
$k=\left\{ \dfrac{\left( y~coordinate~of~A \right)+\left( y~coordinate~of~B \right)}{2} \right\}$
And,
$h=\left\{ \dfrac{\left( \text{a}{{\text{t}}^{2}} \right)+\left( a{{u}^{2}} \right)}{2} \right\}$
$2h=\left( \text{a}{{\text{t}}^{2}} \right)+\left( a{{u}^{2}} \right)$
$\dfrac{2h}{a}=\left\{ {{\text{t}}^{2}}+{{u}^{2}} \right\}...(ii)$
$k=\left\{ \dfrac{\left( 2at \right)+\left( 2au \right)}{2} \right\}$
$k=\left( at \right)+\left( au \right)$
$\dfrac{k}{a}=\left\{ t+u \right\}...(iii)$
So our next task in finding the locus is eliminating the variables from the above equations.
Squaring equation\[\left( iii \right)\]and subtracting it from \[\left( ii \right)\] i.e. \[{{\left( iii \right)}^{2}}-\left( ii \right)\]
Which gives,
${{\left( \dfrac{k}{a} \right)}^{2}}-\dfrac{2h}{a}={{\left( t+u \right)}^{2}}-\left\{ {{\text{t}}^{2}}+{{u}^{2}} \right\}$
${{\left( \dfrac{k}{a} \right)}^{2}}-\dfrac{2h}{a}=\left( {{t}^{2}}+{{u}^{2}}+2ut \right)-\left\{ {{\text{t}}^{2}}+{{u}^{2}} \right\}$
${{\left( \dfrac{k}{a} \right)}^{2}}-\dfrac{2h}{a}=2ut...(iv)$
Squaring equation\[\left( iii \right)\]and subtracting it from \[2\times \left( iv \right)\] i.e. \[{{\left( iii \right)}^{2}}-2\left( iv \right)\]
Which gives,
${{\left\{ \dfrac{k}{a} \right\}}^{2}}-2\left\{ {{\left( \dfrac{k}{a} \right)}^{2}}-\dfrac{2h}{a} \right\}={{\left( t+u \right)}^{2}}-2\left( 2ut \right)$
${{\left( \dfrac{k}{a} \right)}^{2}}-2{{\left( \dfrac{k}{a} \right)}^{2}}+2\dfrac{2h}{a}=\left( {{t}^{2}}+{{u}^{2}}+2ut \right)-4ut$
$-{{\left( \dfrac{k}{a} \right)}^{2}}+\dfrac{4h}{a}=\left( {{t}^{2}}+{{u}^{2}}-2ut \right)$
$\dfrac{4h}{a}-{{\left( \dfrac{k}{a} \right)}^{2}}={{\left( u-t \right)}^{2}}$
$\sqrt{\dfrac{4h}{a}-{{\left( \dfrac{k}{a} \right)}^{2}}}=u-t...(v)$
Substituting equations \[\left( iv \right)\]and \[\left( v \right)\]in \[\left( i \right)\]
\[\tan \alpha \left( 4+\dfrac{{{\left( \dfrac{k}{a} \right)}^{2}}-\dfrac{2h}{a}}{2} \right)=2\sqrt{\dfrac{4h}{a}-{{\left( \dfrac{k}{a} \right)}^{2}}}\]
${{\left( \tan \alpha \right)}^{2}}{{\left\{ 4+\dfrac{{{\left( \dfrac{k}{a} \right)}^{2}}-\dfrac{2h}{a}}{2} \right\}}^{2}}=4\left( \dfrac{4h}{a}-{{\left( \dfrac{k}{a} \right)}^{2}} \right)$
Now since \[\left( h,k \right)\]are general points on our locus we can replace \[h\] by \[x\] and \[k\] by\[y\].
${{\left( \tan \alpha \right)}^{2}}{{\left\{ 4+\dfrac{{{\left( \dfrac{y}{a} \right)}^{2}}-\dfrac{2x}{a}}{2} \right\}}^{2}}=4\left( \dfrac{4x}{a}-{{\left( \dfrac{y}{a} \right)}^{2}} \right)$
This is the required locus.

Note: Students have to think carefully while deciding which is the variable before eliminating. In this question students might eliminate \[~a\] which will give them the wrong answer. Also they may use their different techniques to eliminate the variable from the equations. Also, if they feel it is redundant to use \[\left( h,k \right)\]first and then replace it as \[\left( x,y \right)\] they may use \[\left( x,y \right)\]from the start as well.