Parabola
7.0 Parametric Co-ordinates
7.1 Parametric relation between the coordinates of the ends of a focal chord of parabola
7.2 Important Results
7.0 Parametric Co-ordinates
7.2 Important Results
From the equation of parabola $${y^2} = 4ax$$
We can write it as $$\frac{y}{{2a}} = \frac{{2x}}{y} = t$$
where $'t'$ is a parameter.
Therefore, we get $y=2at$ and $x = a{t^2}$, which are called the parametric equations. The point $(a{t^2},2at)$ is the parametric coordinates.
Similarly, for ${x^2} = 4ay$ type parabola, parametric coordinates are $(2at,a{t^2})$.
Question 5. The equation of parabola is ${y^2} = 4x$. Find the equation of directrix of the locus of a point which divide a line $OP$, as shown in figure $13$ in the ratio $1:3$.
Solution: Let us assume the parametric coordinates of point $P$ on the parabola be $(a{t^2},2at)$ where $'t'$ is a
parameter. From the equation of parabola, $a=1$. Therefore, the coordinates of point $P$ becomes $({t^2},2t)$. Applying section formulae between points $O(0,0)$ and $P({t^2},2t)$, we get the coordinates of point $L(h,k)$ $$h = \frac{{\left( {0 \times 3} \right) + \left( {1 \times {t^2}} \right)}}{{1 + 3}} = \frac{{{t^2}}}{4}$$ or, $${t^2} = 4h$$ and $$k = \frac{{\left( {0 \times 3} \right) + \left( {1 \times 2t} \right)}}{{1 + 3}} = \frac{t}{2}$$ or, $$2t = 4k$$
Therefore, coordinates of point $P$ becomes $(4h,4k)$ which satisfies the equation of parabola i.e., $${\left( {4k} \right)^2} = 4 \times 4h$$ $$16{k^2} = 16h$$ or, $${k^2} = h$$
Therefore, the locus of a point which divide a line $OP$ in the ratio $1:3$ is $${y^2} = x...(1)$$
The equation of directrix is $$x+a=0$$
and from equation $(1)$, $$a = \frac{1}{4}$$
Therefore, the equation of directrix is $$x + \frac{1}{4} = 0$$
