Question

Find the locus of the middle points of chords of the parabola which are of given length \[l\].

Answer 1

Hint: Use distance formula for finding distance between two general points of the parabola.

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 $AB$ is constant i.e. equal to \[l\].
We can write distance $AB$ using distance formula which is,
Distance between two points = $\sqrt{{{\left( x1-x2 \right)}^{2}}-{{\left( y1-y2 \right)}^{2}}}$
So,
$l=\sqrt{{{\left( a{{t}^{2}}-a{{u}^{2}} \right)}^{2}}-{{\left( 2at-2au \right)}^{2}}}$
Squaring both sides we get,
${{l}^{2}}={{a}^{2}}\left\{ {{\left( t-u \right)}^{2}}{{\left( t+u \right)}^{2}} \right\}-4{{a}^{2}}{{\left( t-u \right)}^{2}}...(i)$
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 $AB$.
Using mid-point formula we can write
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 \[\left( iii \right)\]and \[\left( v \right)\] in \[\left( i \right)\] we get,
${{l}^{2}}={{a}^{2}}\left\{ \left( \dfrac{4h}{a}-\left( \dfrac{{{k}^{2}}}{{{a}^{2}}} \right) \right)\left( \dfrac{{{k}^{2}}}{{{a}^{2}}} \right) \right\}-4{{a}^{2}}\left( \dfrac{4h}{a}-\left( \dfrac{{{k}^{2}}}{{{a}^{2}}} \right) \right)$
\[{{l}^{2}}=\left\{ \left( \dfrac{4h}{a}-\left( \dfrac{{{k}^{2}}}{{{a}^{2}}} \right) \right){{k}^{2}} \right\}-4\left( 4ah-{{k}^{2}} \right)\]
\[{{l}^{2}}=\left( \dfrac{4h{{k}^{2}}}{a}-\left( \dfrac{{{k}^{4}}}{{{a}^{2}}} \right) \right)-16ah+4{{k}^{2}}\]
${{l}^{2}}=\dfrac{4h{{k}^{2}}}{a}-\dfrac{{{k}^{4}}}{{{a}^{2}}}-16ah+4{{k}^{2}}$
Now since \[\left( h,k \right)\]are general points on our locus we can replace \[h\] by \[x\] and \[k\] by \[y\].
${{l}^{2}}=\dfrac{4x{{y}^{2}}}{a}-\dfrac{{{y}^{4}}}{{{a}^{2}}}-16ax+4{{y}^{2}}$
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.