Question

The locus of the middle points of the chords of the parabola ${{y}^{2}}=4ax$ which passes
through the origin is
(a) ${{y}^{2}}=ax$
(b) ${{y}^{2}}=2ax$
(c) ${{y}^{2}}=4ax$
(d) ${{x}^{2}}=4ay$

Answer 1

Hint: Consider any two points on parabola with parameter $t$ and write the equation of chord joining them. Pass the equation through origin and solve them to find the locus of the point joining middle points of chord.

We have the parabola ${{y}^{2}}=4ax$. To find the locus of middle point of the chords
joining two points on the parabola, we will assume two points on the parabola of the
form $P({{t}_{1}})=\left( at_{1}^{2},2a{{t}_{1}} \right)$and $Q({{t}_{2}})=\left( at_{2}^{2},2a{{t}_{2}} \right)$.
We know that the equation of chords joining two points $P({{t}_{1}})$and$Q({{t}_{2}})$ on the
parabola is $y\left( {{t}_{1}}+{{t}_{2}} \right)=2x+2a{{t}_{1}}{{t}_{2}}$.
We know that this chord passes through the origin. So, we will substitute the point$\left( 0,0 \right)$in the equation of the chord.

Substituting the point$\left( 0,0 \right)$in the equation of chord, we get$0\left( {{t}_{1}}+{{t}_{2}} \right)=2\times 0+2a{{t}_{1}}{{t}_{2}}$.
Hence, let’s assume${{t}_{2}}=0$.
We observe that any chord of the parabola which is passing through origin has origin as one of its
end points.
We can assume other end of the chord to be $P({{t}_{1}})=\left( at_{1}^{2},2a{{t}_{1}} \right)$.
To find the middle point of two points of the form $\left( {{x}_{1}},{{y}_{1}} \right)$and$\left( {{x}_{2}},{{y}_{2}} \right)$, use the formula $\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$.
Substituting${{x}_{1}}=0,{{y}_{1}}=0,{{x}_{2}}=at_{1}^{2},{{y}_{2}}=2a{{t}_{1}}$in the above formula,
we get the middle point of $\left( 0,0 \right)$and$P({{t}_{1}})=\left( at_{1}^{2},2a{{t}_{1}} \right)$as$\left( \dfrac{0+at_{1}^{2}}{2},\dfrac{0+2a{{t}_{1}}}{2} \right)$.
Thus, the middle point of chords has the form $\left( \dfrac{at_{1}^{2}}{2},a{{t}_{1}} \right)$.
To find the locus of middle point of the chords, let’s assume $x=\dfrac{at_{1}^{2}}{2},y=a{{t}_{1}}$.
Eliminating ${{t}_{1}}$from both equations by rearranging the terms, we
get $\dfrac{2x}{a}=t_{1}^{2},\dfrac{y}{a}={{t}_{1}}$.
Substituting the value of ${{t}_{1}}$from both equations, we get $\dfrac{2x}{a}={{\left( \dfrac{y}{a} \right)}^{2}}$.
Rearranging the terms, we get ${{y}^{2}}=2ax$.
Hence, the correct answer is ${{y}^{2}}=2ax$.

Note: It’s very necessary to use the fact that any chord of the parabola passing through origin has
origin as one of its end points. We also verified this fact by assuming any two points on the parabola
and making the equation of chord pass through origin.