Question

The normal at the ends of the latus rectum of the parabola ${{y}^{2}}=4x$ meet the parabola again at A and A’, then find the length of AA’.

Answer 1

To solve this problem we need to know that the ends of the latus rectum of a parabola ${{y}^{2}}=4ax$ are given by (a, 2a) and (a, -2a). and also if the normal drawn to the parabola ${{y}^{2}}=4ax$ at a parameter ${{t}_{1}}$ then it again meets the same parabola at the parameter $t=-{{t}_{1}}-\dfrac{2}{{{t}_{1}}}$. So using the above properties we will first find the parameter of the latus rectum of the given parabola and from that we will find the new parameters at which the normal at the latus rectum of the parabola intersects the parabola again. After finding the parameter we will find the coordinates of that points and then using distance formula will find the length of the line joining them.

Complete step-by-step answer:
We are given the parabola whose equation is given as,
${{y}^{2}}=4x$
And also it is mentioned in the question that the normal drawn at the both ends of the parabola again intersects the parabola at A and A’ then we have to find the length AA’.
We know that if we have the parabola ${{y}^{2}}=4ax$ then it’s coordinates of the length of the latus rectum are given by (a, 2a) and (a, -2a) and also if normal drawn to this parabola at $\left( a{{t}^{2}},2at \right)$ where t is the parameter of that point then that normal again intersects the parabola with the parameter of that point given by ${{t}_{1}}=-t-\dfrac{2}{t}$.
So now we have parabola as,
${{y}^{2}}=4x$
By comparing it with the parabola ${{y}^{2}}=4ax$we get,
a = 1,
so coordinates of the ends of the latus rectum are,
= (a, 2a) and (a, -2a)
Putting a = 1, we get
= (1, 2) and (1, -2)
Now if we consider the point (1, 2) its parameter is given by, (by comparing with general y coordinate)
$\begin{aligned} & 2t=2 \\\ & t=1 \\\ \end{aligned}$
So the parameter where normal at this point again intersects the parabola i.e. at point A,
$\begin{aligned} & t'=-1-\dfrac{2}{1} \\\ & t'=-3 \\\ \end{aligned}$
Coordinates of the point A where it intersect the parabola we get as,
$\begin{aligned} & =\left( {{t}^{2}},2t \right) \\\ & =\left( {{\left( 3 \right)}^{2}},2\times -3 \right) \\\ & =\left( 9,-6 \right) \\\ \end{aligned}$
Similarly for the other end of the latus rectum with coordinates as (1, -2),
Its parameter is given by, (by comparing with general y coordinate)
$\begin{aligned} & 2t=-2 \\\ & t=-1 \\\ \end{aligned}$
So the parameter where normal at this point again intersects the parabola i.e. at point A’,
$\begin{aligned} & t'=-\left( -1 \right)-\dfrac{2}{\left( -1 \right)} \\\ & t'=1+2 \\\ & t'=3 \\\ \end{aligned}$
Coordinates of the point A’ where it intersect the parabola we get as,
$\begin{aligned} & =\left( {{t}^{2}},2t \right) \\\ & =\left( {{\left( 3 \right)}^{2}},2\times 3 \right) \\\ & =\left( 9,6 \right) \\\ \end{aligned}$
Now to find the length AA’ we will use the distance formula on the points A(9, -6) and A’(9, 6),

& =\sqrt{{{\left( 9-9 \right)}^{2}}+{{\left( 6-\left( -6 \right) \right)}^{2}}} \\\ & =\sqrt{{{\left( 12 \right)}^{2}}} \\\ & =12 \\\ \end{aligned}$$ Hence we get the length as 12 units.  **Note:** To solve this problem you need to have prior knowledge that how to find the coordinates of the point where normal again intersects the parabola and also you have to know how to find the coordinates of the latus rectum of the parabola and what is meant by it. In the above question we could have also found the parameter of the obtained coordinates of the latus rectum by comparing it with the x-coordinate instead of the y-coordinate.