Condition for a Chord to Pass Through the Focus

The standard form of the parabola is

$$y^2=4ax$$

Hence

$$4a=12$$

which gives

$$a=3$$

Let the points $$P$$ and $$Q$$ correspond to parameters $$t_1$$ and $$t_2$$ respectively.

Since

$$PQ$$

is a focal chord, we have

$$t_1t_2=-1$$

The length of a focal chord is

$$L=a(t_1-t_2)^2$$

Given $$L=48$$ and $$a=3$$

Therefore

$$48=3(t_1-t_2)^2$$

$$ (t_1-t_2)^2=16 $$

$$ |t_1-t_2|=4 $$

Now use the standard result that the area of the triangle formed by the tangents at the ends of a focal chord and the focal chord itself is

$$\text{Area}=\frac{a^2}{2}|t_1-t_2|^3$$

Substituting

$$a=3$$ and $$|t_1-t_2|=4$$ gives

$$\text{Area}=\frac{3^2}{2}(4)^3$$

$$=\frac{9}{2}\cdot64$$

$$=9\cdot32$$

$$=288$$

Final Answer :

$$288$$