Question

Find the locus of the middle points of chords of the parabola which pass through the focus.

Answer 1

Hint: We have to consider the endpoints of the chord of the parabola as $A\left( a{{t}^{2}},2at \right)$ and $B\left( a{{t}^{2}}, -2at \right)$. Then using the midpoint formula, we can find the midpoint of the chord of the parabola. Now to find its locus, we have to manipulate the midpoints obtained to get an equation.

Complete step-by-step answer:

We know that points on a standard parabola $\left( {{y}^{2}}=4ax \right)$ can be parameterized by a parameter (say t) i.e. it can be written as $\left( a{{t}^{2}},2at \right)$, where the point changes as t changes.
If one endpoint of the parabola is $A\left( a{{t}^{2}},2at \right)$, then the other endpoint will be $B\left( a{{t}^{2}}, -2at \right)$.
We can check this by writing the equation of the focal chord using the two-point form of a line since we know that point $A$ and focus $\left( a,0 \right)$ lie on the chord. Then we just have to check where the chord intersects the parabola again, which can be done by solving the equation of the focal chord and parabola together.
In order to find the locus of the point, let us consider a point with coordinates $\left( h,k \right)$ which lies on the locus. Since $\left( h,k \right)$ lies on the locus, which is the locus of mid-points of the focal chords of the parabola, we can use the midpoint formula.
Using midpoint formula we can write that ${{x}_{mid}}=\dfrac{{{x}_{1}}+{{x}_{2}}}{2}$ and ${{y}_{mid}}=\dfrac{{{y}_{1}}+{{y}_{2}}}{2}$. Therefore, we get
h=\left\\{ \dfrac{\left( x~coordinate~of~A \right)+\left( x~coordinate~of~B \right)}{2} \right\\}
h=\left\\{ \dfrac{\left( a{{t}^{2}} \right)+\left( \dfrac{a}{{{\text{t}}^{2}}} \right)}{2} \right\\}
$2h=\left( a{{t}^{2}} \right)+\left( \dfrac{a}{{{\text{t}}^{2}}} \right)$
\dfrac{2h}{a}=\left\\{ \left( {{t}^{2}} \right)+\left( \dfrac{1}{{{\text{t}}^{2}}} \right) \right\\}...(i)
k=\left\\{ \dfrac{\left( y~coordinate~of~A \right)+\left( y~coordinate~of~B \right)}{2} \right\\}
k=\left\\{ \dfrac{\left( 2at \right)+\left( -\dfrac{2a}{t} \right)}{2} \right\\}
$2k=\left( 2at \right)+\left( -\dfrac{2a}{t} \right)$
\dfrac{k}{a}=\left\\{ \left( t \right)-\left( \dfrac{1}{t} \right) \right\\}...(ii)
Now we have to combine the above two equations and remove all variables. Only $t$ is the variable here. Squaring equation $\left( ii \right)$ and subtracting it from $\left( i \right)$ i.e. ${{\left( ii \right)}^{2}}-\left( i \right)$, we get,
{{\left( \dfrac{k}{a} \right)}^{2}}-\dfrac{2h}{a}={{\left\\{ \left( t \right)-\left( \dfrac{1}{t} \right) \right\\}}^{2}}-\left\\{ \left( {{t}^{2}} \right)+\left( \dfrac{1}{{{\text{t}}^{2}}} \right) \right\\}
{{\left( \dfrac{k}{a} \right)}^{2}}-\dfrac{2h}{a}=\left\\{ {{\left( t \right)}^{2}}+{{\left( \dfrac{1}{t} \right)}^{2}}-2\left( t \right)\left( \dfrac{1}{t} \right) \right\\}-\left\\{ \left( {{t}^{2}} \right)+\left( \dfrac{1}{{{\text{t}}^{2}}} \right) \right\\}
${{\left( \dfrac{k}{a} \right)}^{2}}-\dfrac{2h}{a}=-2$
Now since $\left( h,k \right)$ are general points on our locus we can replace $h$ by $x$ and $k$ by $y$. Therefore, we get
${{\left( \dfrac{y}{a} \right)}^{2}}-\dfrac{2x}{a}=-2$
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.