Question

If double ordinate of parabola \[{y^2} = 4ax\;\] subtends right angle at vertex, then length of double ordinate is double the lactic rectum \[\left( {8a} \right)\].

Answer 1

Hint:
The equation of the curve shows us that it is a parabola. The double ordinate is the same in length as the latus rectum of the parabola. Hence, it is the latus rectum.
We will prove the property in reverse by assuming the lactic rectum is \[\left( {8a} \right)\] and proving double ordinate subtends right angle at vertex.
Use this information to find the ordinates of the two ends of the latus rectum as 4a and -4a. Put these values in the equation of the curve to get the x-coordinate, and thus the coordinates of the end points. The vertex of the curve is the origin. Find the slope of the two lines connecting the origin to the end points. The product of these slopes will define that the two lines are perpendicular to each other.

Complete step by step solution:
We can observe that the given equation of the curve is of the form \[{y^2} = 4ax\;\], which is the equation of a parabola. Hence, the given curve is a parabola.
We will prove the property in reverse by assuming the lactic rectum is\[\left( {8a} \right)\]and proving double ordinate subtends right angle at vertex.
We will assume that the length of the latus rectum of a parabola having equation \[{y^2} = 4ax\;\] is 8a. Hence, the double ordinate given to us is a latus rectum of the parabola. Also, we can observe from the equation of the parabola that the vertex is at the origin $\left( {0,0} \right)$.
Thus, the length of the latus rectum is 8a. So, the length of the semi latus rectum will be 4a.
This gives us the fact that the coordinates of the end points of the latus rectum can be given as $\left( {x,4a} \right)$and $\left( {x, - 4a} \right)$.
These points lie on the parabola and hence, satisfy the equation of the parabola. Thus, putting these values in the equation of the parabola we get
${\left( { \pm 4a} \right)^2} = 4ax$
$ \Rightarrow 16{a^2} = 4ax$
$ \Rightarrow x = \dfrac{{16{a^2}}}{{4a}}$
$ \Rightarrow x = 4a$
Thus, the end points of the latus rectum come out to be $\left( {4a,4a} \right)$ and $\left( {4a, - 4a} \right)$.
Let us denote these points as $P\left( {4a,4a} \right)$ and $Q\left( {4a, - 4a} \right)$.
We now find the slope of lines OP and OQ, where O is the origin.
Slope of line OP is given as ${m_1} = \dfrac{{4a - 0}}{{4a - 0}} = 1$
Slope of line OQ is given as ${m_2} = \dfrac{{ - 4a - 0}}{{4a - 0}} = - 1$
The product of these two slopes ${m_1}{m_2} = - 1$.
This is possible only if the two lines are perpendicular to each other.
Thus, the lines OP and OQ are perpendicular to each other.

Hence proved that if double ordinate of parabola \[{y^2} = 4ax\;\] subtends right angle at vertex, then length of double ordinate is double the lactic rectum \[\left( {8a} \right)\].

Note:
It must be observed that the situation given in the question, where the double ordinate is of length 8a is a special case and the result is valid only in this case. The angle between the line’s changes with change in the length of the double ordinate. Also, it is an important result and advisable to be memorized, that the lines joining the vertex to the end points of the latus rectum of a parabola are perpendicular to each other.