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.