Question

Prove that the locus of the mid-point of all tangents drawn from the point on the directrix to the parabola \[{{y}^{2}}=4ax\] is \[{{y}^{2}}(2x+a)=a{{(3x+a)}^{2}}\].

Answer 1

Hint: Write the equation of tangent at any point on the parabola and substitute any point on the directrix in this equation of tangent. Get the exact point on the directrix which satisfies the equation of tangent and then find the mid-point of the two points.

Complete step-by-step answer:

Consider a parabola \[{{y}^{2}}=4ax\] . Its vertex is at the point \[O\left( 0,0 \right)\]. We want to find the locus of mid-point of all tangents drawn from a point on the directrix to the parabola.
We know that the equation of directrix of the parabola \[{{y}^{2}}=4ax\] is \[x=-a\].

Consider any point on the parabola \[{{y}^{2}}=4ax\] of the form \[D\left( a{{t}^{2}},2at \right)\] such that the tangent from a point \[A\left( x,y \right)\] on the directrix touches the parabola at \[D\left( a{{t}^{2}},2at \right)\].

We know that the equation of tangent to the parabola \[{{y}^{2}}=4ax\] at any point \[D\left( a{{t}^{2}},2at \right)\] is of the form \[yt=x+a{{t}^{2}}\].

The point \[A\left( x,y \right)\] lies on the directrix \[x=-a\] of the parabola. Hence, we can write the point \[A\left( x,y \right)\] as \[A\left( -a,y \right)\].

This point \[A\left( -a,y \right)\] also lies on the tangent of the parabola \[yt=x+a{{t}^{2}}\].
Substituting the point in the equation of tangent, we get \[ty=-a+a{{t}^{2}}\].

Thus, we have \[y=\dfrac{-a}{t}+at\].

Hence, the coordinates of point \[A\left( -a,y \right)\] are \[A\left( -a,\dfrac{-a}{t}+at \right)\].
We now want to find the mid-point of the points \[A\left( -a,\dfrac{-a}{t}+at \right)\] and \[D\left( a{{t}^{2}},2at \right)\].

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

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

We want to find the locus of this mid-point.
Let’s assume \[\left( x,y \right)\] is the locus of the point \[\left( \dfrac{-a+a{{t}^{2}}}{2},\dfrac{\dfrac{-a}{t}+3at}{2} \right)\].

Thus, we have \[x=\dfrac{-a+a{{t}^{2}}}{2},y=\dfrac{\dfrac{-a}{t}+3at}{2}\].

Solving the above equation \[x=\dfrac{-a+a{{t}^{2}}}{2}\] by rearranging the terms, we get \[{{t}^{2}}=\dfrac{2x+a}{a}\].

Hence, we have \[t=\sqrt{\dfrac{2x+a}{a}}\].

Substituting the above equation in the equation \[y=\dfrac{\dfrac{-a}{t}+3at}{2}\], we get \[2y=\dfrac{-a}{\sqrt{\dfrac{2x+a}{a}}}+3a\sqrt{\dfrac{2x+a}{a}}\]

Rearranging the terms, we get \[2y\sqrt{\dfrac{2x+a}{a}}=-a+3a\left( \dfrac{2x+a}{a} \right)\]
On further solving, we get \[2y\sqrt{\dfrac{2x+a}{a}}=-a+6x+3a=2a+6x\]
\[\begin{align}
  & \Rightarrow 2y\sqrt{\dfrac{2x+a}{a}}=2\left( a+3x \right) \\
 & \Rightarrow y\sqrt{\dfrac{2x+a}{a}}=\left( a+3x \right) \\
\end{align}\]

Squaring the terms on both sides, we get \[{{y}^{2}}\left( \dfrac{2x+a}{a} \right)={{\left( a+3x \right)}^{2}}\]
\[\Rightarrow {{y}^{2}}\left( 2x+a \right)=a{{\left( a+3x \right)}^{2}}\]

Hence, the locus of mid-point of tangent from directrix to the parabola is \[{{y}^{2}}\left( 2x+a \right)=a{{\left( a+3x \right)}^{2}}\]

Note: We can also solve this question by writing the equation of tangent from a point on the directrix (considering it as a point outside the parabola) and then finding its point of intersection with parabola.