It is a fundamental concept. The end points of latus rectum of the parabola $x^2=4ay$ are (-2a , a) , (2a, a).

a = 4 , vertex = (0,0) , focus = (0,-4) . 

$y^2=4.\frac{1}{5}x;a=\frac{1}{5}$ . Focus is $(\frac{1}{5},0)$ and co-ordinates of latus rectum are $y^2=\frac{4}{25}\Rightarrow y=\pm \frac{2}{5}$ 

or end points of latus rectum are $(\frac{1}{5},\pm \frac{2}{5})$ . 

$x^2=-8y\Rightarrow a=-2.$ So , focus = (0,-2)

Ends of latus rectum = (4,-2) , (-4,-2) .

Trick: Since the ends of latus rectum lie on parabola , so only points (-4,-2) and (4,-2) satsify the parabola. 

a = 3 $\Rightarrow$ abscissa is 4 - 3 = 1 and $y^2=12,y=\pm 2\sqrt{3}.$ 

Hence points are $(1,2\sqrt{3}),(1,-2\sqrt{3})$ . 

$y_1=3x_1$ . According to given condition $9x_1^2=36x_1$ 

$\Rightarrow x_1=4,0\Rightarrow y_1=12,0$ 

Hence the points are (0,0) and (4,12). 

Distance between focus and directrix is 

= $\left|\frac{3-4-2}{\sqrt{2}}\right|=\frac{\pm 3}{\sqrt{2}}$ 

Hence latus rectum = $3\sqrt{2}$ 

( Since latus rectum is two times the distance between focus and directrix ) . 

Required locus is $(3y{)}^2$ = 4ax 

$\Rightarrow$ $9y^2=4ax$ . 

Let the equation of parabola is $x^2=4ay,$ but a = $\frac{4}{-2}=-2.Thenequationis{x}^2=-8y$ and latus rectum = 4a = 8 

The point (-3,2) will satisfy the equation $y^2=4ax$ 

$\Rightarrow$ 4 = -12a $\Rightarrow 4a=-\frac{4}{3}=\frac{4}{3},$ (Taking positive sign).