10.1 Point form

  Parabola

Question 7. A normal and tangent is drawn at a point $P$ on the parabola ${y^2} = 4ax$ which cuts the $X$-axis at $N$ and $T$ as shown in figure $23$. The locus of centroid of $\Delta PTN$ is a parabola whose

(i) Vertex is $(\frac{{2a}}{3},0)$

(ii) Directrix is $x=0$

(iii) Latus rectum is $\frac{{2a}}{3}$

(iv) Focus is $(a,0)

Solution: Let us assume the coordinates of point $P$ be $(a{t^2},2at)$. Therefore, the equation of normal is $$y - 2at = - t\left( {x - a{t^2}} \right)$$

Put $y=0$, we get $$x = 2a + a{t^2}$$

The coordinates of point $N$ is $(2a + a{t^2},0)$.

Similarly, the equation of tangent is $$y - 2at = \frac{1}{t}\left( {x - a{t^2}} \right)$$

Put $y=0$, we get $$x = - a{t^2}$$

The coordinates of point $T$ is $( - a{t^2},0)$.

Let us assume the coordinates of centroid of $\Delta PTN$ be $(h,k)$. Therefore,

$$h = \frac{{a{t^2} + 2a + a{t^2} - a{t^2}}}{3} = \frac{{2a + a{t^2}}}{3}...(1)$$ and

$$k = \frac{{0 + 2at + 0}}{3} = \frac{{2at}}{3}...(2)$$or, $$t = \frac{{3k}}{{2a}}$$

Put the value of $t$ in equation $(1)$ , we get $$3h = 2a + a \times {(\frac{{3k}}{{2a}})^2}$$ $$3h = 2a + \frac{{9{k^2}}}{{4a}}$$ $$3h = \frac{{8{a^2} + 9{k^2}}}{{4a}}$$

Therefore, the locus of centroid is $$12ax = 9{y^2} + 8{a^2}$$

or, $${y^2} = \frac{{4a}}{9}\left( {3x - 2a} \right)...(3)$$

which is the equation of parabola.

Now, Let us assume $Y=y$ and $X=3x-2a$

Equation $(3)$ becomes $${Y^2} = \frac{{4a}}{9}X...(4)$$

The vertex of parabola given in equation $(4)$ is $(0,0)$. Therefore, the vertex of parabola given in equation $(3)$ can be find out by putting $X=0$ and $Y=0$ i.e., $$3x-2a=0$$ $$x = \frac{{2a}}{3}$$ and $$y=0$$

So, the vertex of parabola given in equation $(3)$ is $V(\frac{{2a}}{3},0)$.

The focus of parabola given in equation $(4)$ is $(\frac{a}{9},0)$ and the focus of parabola given in equation $(3)$ can be find out by putting $X = \frac{a}{9}$ and $Y=0$ i.e., $$3x - 2a = \frac{a}{9}$$

or, $$x = \frac{{19a}}{{27}}$$ and $$y=0$$

Therefore, the focus of parabola given in equation $(3)$ is $(\frac{{19a}}{{27}},0)$.

Equation of directrix of parabola given in equation $(4)$ is $$X + \frac{a}{9} = 0$$ or, $$3x - 2a + \frac{a}{9} = 0$$ $$x = \frac{{17a}}{{27}}$$

Length of latus rectum$ = \frac{{4a}}{9}$

Therefore, correct option is $(i)$.