Question

PQ is a double ordinate of a parabola. Find the locus of its point of intersection.

Answer 1

Hint: If a point divides a line in $m:n$ internally then coordinates of a point are given by $p(x)=\dfrac{m({{x}_{1}})+n({{x}_{2}})}{m+n}$ and $p(y)=\dfrac{m({{y}_{1}})+n({{y}_{2}})}{m+n}$. The equation of a parabola is given by ${{y}^{2}}=4ax$ for right-handed parabola and ${{y}^{2}}=-4ax$ left-handed parabola. In question, it is not mentioned which parabola it is, therefore, we can take the most general case i.e. right-handed parabola.

Complete step by step answer:
We have been given that PQ is a double ordinate of a parabola. Here, we have to find the locus of its point of trisection.
Before proceeding with the question, we must know that the double-ordinate is the chord of a parabola which is perpendicular to the axis.
Let PQ be the double-ordinate of the parabola and $S(h,k)$ be the point which trisects the double ordinate PQ of the parabola. As $S(h,k)$ trisects the double ordinate PQ, it means it divides the double ordinate in the ratio $1:2$ . The figure is as below.

As we know that the general coordinates passing through the parabola is given by $(a{{t}^{2}},2at)$ and $(a{{t}^{2}},-2at)$, we know that the endpoints of the double ordinate pass through the parabola. Therefore, we can take P as $(a{{t}^{2}},2at)$ and Q as $(a{{t}^{2}},-2at)$.
Therefore, according to the section formula, $S(x)=\dfrac{m({{x}_{1}})+n({{x}_{2}})}{m+n}$ we get the $x$ and $y$ coordinates of the point of trisection of double-ordinate as follows:
$S(x)=\dfrac{1(a{{t}^{2}})+2(a{{t}^{2}})}{1+2}$
$\Rightarrow S(x)=\dfrac{3a{{t}^{2}}}{3}$
$\therefore S(x)=a{{t}^{2}}.....(i)$
$S(y)=\dfrac{1(2at)+2(-2at)}{1+2}$
$\Rightarrow S(y)=\dfrac{-2at}{3}.....(ii)$
We have $S(h,k)$ as the coordinates of the point of trisection. Therefore we can replace $S\left( x \right)$ as $h$ from equation (i), we get,
$\therefore h=a{{t}^{2}}$
$\therefore h=a{{t}^{2}}...(iii)$
After replacing $S\left( y \right)$ as $k$ equation (ii) we get,
$k=\dfrac{-2at}{3}$
$\Rightarrow t=\dfrac{-3k}{2a}$
Substitute the value of $t$ in equation (iii) we get:
$\Rightarrow h=a{{\left( \dfrac{-3k}{2a} \right)}^{2}}$
$\Rightarrow h=a\left( \dfrac{9{{k}^{2}}}{4{{a}^{2}}} \right)$
\[\Rightarrow 9{{k}^{2}}a=4{{a}^{2}}h\]
$\therefore 9{{k}^{2}}=4ah$
Replacing $h$ by $x$ and $k$ by $y$ we get:
$9{{y}^{2}}=4ax$
Hence, $9{{y}^{2}}=4ax$ is the required locus.

Note: Do not confuse double ordinate as latus rectum of the parabola, the double ordinate is any chord which is perpendicular to parabola and latus rectum is the double ordinate of the parabola which passes through the focus of the parabola. In any question, if you are asked to find the locus of the parabola, then try to assume general points on the parabola and then resolve it into $x$ and $y$. When you are not given whether the parabola is right-handed or left-handed, in that case, use the most general form i.e. ${{y}^{2}}=4ax$.