Question

Prove that the locus of the mid-points of the chord of a parabola passes through the vertex is a parabola.

Answer 1

Hint: Take any general point on the parabola and find the mid-point of the chord with the end points being the vertex and one point on the parabola. Solve the equations to eliminate the parameters and get the locus of mid-point of chord of the parabola which passes through the vertex of the parabola.

Complete step-by-step answer:

Consider a parabola \[{{y}^{2}}=4ax\]. We want to prove that the middle point of chord of the parabola which passes through the vertex is a parabola.

We know that the vertex of the parabola of the form \[{{y}^{2}}=4ax\] is \[O\left( 0,0 \right)\].
Let us consider any point on the parabola\[{{y}^{2}}=4ax\] of the form to be \[A\left( a{{t}^{2}},2at \right)\].
Let us consider an arbitrary chord \[OA\] of the parabola which passes through the vertex \[O\left( 0,0 \right)\] and \[A\left( a{{t}^{2}},2at \right)\] of the parabola.

We want to find the mid-point of this chord of the parabola.

We know that the mid-point of any two points of the form \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] is \[\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)\].

Substituting \[{{x}_{1}}=0,{{y}_{1}}=0,{{x}_{2}}=a{{t}^{2}},{{y}_{2}}=2at\] in the above equation, we get \[\left( \dfrac{0+a{{t}^{2}}}{2},\dfrac{0+2at}{2} \right)\] as the mid-point of the chord joining the points \[O\left( 0,0 \right)\] and \[A\left( a{{t}^{2}},2at \right)\].

Let’s assume that the locus of mid-point of any chord of the parabola passing through vertex is of the form \[\left( x,y \right)\].
Thus, we have \[\left( x,y \right)=\left( \dfrac{0+a{{t}^{2}}}{2},\dfrac{0+2at}{2} \right)\].

Comparing the terms, we get \[x=\dfrac{a{{t}^{2}}}{2},y=at\].

Rearranging the terms in both equations in terms of the parameter \[t\], we get \[\dfrac{2x}{a}={{t}^{2}},\dfrac{y}{a}=t\].

Substituting one equation to the other one, we get \[\dfrac{2x}{a}={{\left( \dfrac{y}{a} \right)}^{2}}\].
Simplifying the terms, we get \[\dfrac{2x}{a}={{\left( \dfrac{y}{a} \right)}^{2}}=\dfrac{{{y}^{2}}}{{{a}^{2}}}\].

Thus, we have \[{{y}^{2}}=2ax\].

Hence, we get \[{{y}^{2}}=2ax\] as the locus of mid-point of chord of the parabola which passes through the vertex of the parabola.

Note: We can take any point on the parabola of the form \[\left( h,k \right)\] and solve the equations to find the mid-point of the chord of the parabola passing through the vertex.