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.