Given parabola $${x^2} = 12y$$   is which is of the form $${x^2} = 4ay$$
$$ \Rightarrow 4a = 12 \Rightarrow a = 3$$
Now, $$LM$$  is the latus rectum whose length $$ = 4a = 4 \times 3 = 12$$
So, area of $$\Delta LMV$$
$$\eqalign{ & = \frac{1}{2} \times LM \times VF \cr & = \left( {\frac{1}{2} \times 12 \times 3} \right){\text{square units}} \cr & = 18{\text{ square units}} \cr} $$

The parabolas are $${y^2} - x = 0$$   and $${y^2} + x = 0.$$   The point $$\left( {\lambda ,\, - 1} \right)$$  is an exterior point if $$1 - \lambda > 0$$   and $$1 + \lambda > 0\,\,\, \Rightarrow \lambda < 1$$     and $$\lambda > - 1\,\,\, \Rightarrow - 1 < \lambda < 1.$$

Vertex of a parabola is the mid point of focus and the point where directrix meets the axis of the parabola.
Here focus is $$O\left( {0,\,0} \right)$$  and directrix meets the axis at $$B\left( {2,\,0} \right)$$
$$\therefore $$ Vertex of the parabola is $$\left( {1,\,0} \right).$$

$$y=mx+c$$   is normal to the parabola
$${y^2} = 4ax$$   if $$c = - 2am - a{m^3}$$
Here $$m=-1, \,c=k$$     and $$a=3$$
$$\therefore c = k = - 2\left( 3 \right)\left( { - 1} \right) - 3{\left( { - 1} \right)^3} = 9$$

$$\left( {\lambda ,\,\mu } \right)$$  is interior to both the curves if $${\lambda ^2} + {\mu ^2} - 16 < 0$$    and $${\mu ^2} - 4\lambda < 0$$
Now, $${\mu ^2} - 4\lambda < 0\,\,\,\, \Rightarrow \lambda > {\left( {\frac{\mu }{2}} \right)^2}$$
Hence, if $$\mu = 0,\,\lambda = 1,\,2,\,3,\,.....;$$     if $$\mu = 1,\,\lambda = 1,\,2,\,3,\,.....;$$     if $$\mu = 2,\,\lambda = 2,\,3,\,.....;$$     if $$\mu = 3,\,\lambda = 3,\,4\,.....;$$
Also $${\lambda ^2} + {m^2} - 16 < 0\,\,\,\,\, \Rightarrow {\lambda ^2} < 16 - {\mu ^2}$$
Hence, if $$\mu = 0,\,\lambda = 1,\,2,\,3\,;$$    if $$\mu = 1,\,\lambda = 1,\,2,\,3\,;$$    if $$\mu = 2,\,\lambda = 2,\,3\,;$$    if $$\mu = 3,\,\lambda $$  has no integral value.
$$\therefore \,\,\left( {1,\,0} \right),\,\left( {2,\,0} \right),\,\left( {3,\,0} \right),\,\left( {1,\,1} \right),\,\left( {2,\,1} \right),\,\left( {3,\,1} \right),\,\left( {2,\,2} \right),\,\left( {3,\,2} \right)$$           are the possible points.

Let $$P\left( {h,\,k} \right)$$  divides $$OQ$$  in the ratio $$1 : 3$$
Let any point $$Q$$ on $${x^2} = 8y$$   is $$\left( {4t,\,2{t^2}} \right).$$
Then by section formula
$$\eqalign{ & \Rightarrow k = \frac{{{t^2}}}{2}{\text{ and }}h = t \cr & \Rightarrow 2k = {h^2} \cr} $$
Required locus of $$P$$ is $${x^2} = 2y$$

The given curves are
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,{y^2} = 8x.....(1) \cr & {\text{and }}xy = - 1.....(2) \cr} $$
If $$m$$ is the slope of tangent to (1), then equation of tangent is
$$y = mx + \frac{2}{m}$$
If this tangent is also a tangent to (2), then putting value of $$y$$ in curve (2)
$$\eqalign{ & x = \left( {mx + \frac{2}{m}} \right) = - 1 \cr & \Rightarrow m{x^2} + \frac{2}{m}x + 1 = 0 \cr & \Rightarrow {m^2}{x^2} + 2x + m = 0 \cr} $$
We should get repeated roots for the equation (condition of tangency)
$$\eqalign{ & \Rightarrow D = 0 \cr & \therefore {\left( 2 \right)^2} - 4{m^2}.m = 0 \cr & \Rightarrow {m^3} = 1\,\,\,\,\, \Rightarrow m = 1 \cr} $$
Hence required tangent is $$y=x +2$$

The tangents to the parabola $${y^2} = 4ax$$   at the points $$\left( {a,\,2a} \right),\,\left( {a,\, - 2a} \right)$$    are $$y = x + a$$   and $$y = - x - a.$$    The third side of the triangle is $$x = a.$$  Clearly, these lines form a right-angled triangle whose two sides are equal.

Since the axis is horizontal and vertex is $$\left( { - 3,\, - 2} \right),$$
$$\therefore $$  the equation of the parabola must be of the form $${\left( {y + 2} \right)^2} = 4a\left( {x + 3} \right)$$
It passes through $$\left( {1,\,2} \right),$$  so $$16 = 16a$$   i.e $$a = 1.$$
Hence, the equation of the required parabola is
$${\left( {y + 2} \right)^2} = 4\left( {x + 3} \right){\text{ or }}{y^2} + 4y - 4x - 8 = 0$$

KEY CONCEPT : The directrix of the parabola $${y^2} = 4a$$
$$\left( {x - {x_1}} \right)$$  is given by $$x = {x_1} - a$$
$${y^2} = kx - 8 \Rightarrow {y^2} = k\left( {x - \frac{8}{k}} \right)$$
Directrix of parabola is $$x = \frac{8}{k} - \frac{k}{4};$$
Now, $$x=1$$  also coincides with $$x = \frac{8}{k} - \frac{k}{4}$$
On comparison, $$\frac{8}{k} - \frac{k}{4} = 1,$$   or $${k^2} - 4k - 32 = 0$$
On solving we get $$k=4$$