Given parabola
$$y^2=16x$$
Comparing with
$$y^2=4ax$$
we get
$$a=4$$
Hence the focus is
$$S=(4,0)$$
Now find the tangent to the parabola at
$$P(16,16)$$
The tangent to $$y^2=16x$$ at $$(x_1,y_1)$$ is $$yy_1=8(x+x_1)$$
Substituting
$$P=(16,16)$$
$$16y=8(x+16)$$
$$x-2y+16=0$$
Therefore the normal at $$P$$ is $$y+2x-48=0$$ or $$y=48-2x$$
Let the centre of the circle be $$O(h,k)$$
Since the circle touches the parabola at $$P$$,
the radius $$OP$$ lies along the normal.
Hence
$$O=(h,48-2h)$$
The circle passes through both $$P$$ and $$S$$
Therefore
$$OP=OS$$
Using distance formula,
$$ (h-16)^2+(48-2h-16)^2=(h-4)^2+(48-2h)^2 $$
$$ (h-16)^2+(32-2h)^2=(h-4)^2+(48-2h)^2 $$
Expanding,
$$h^2-32h+256+4h^2-128h+1024=h^2-8h+16+4h^2-192h+2304$$
$$5h^2-160h+1280=5h^2-200h+2320$$
$$40h=1040$$
$$h=26$$
Hence
$$k=48-2(26)=-4$$
Therefore the centre is
$$O=(26,-4)$$
Radius squared is
$$r^2=(26-4)^2+(-4)^2$$
$$=22^2+16$$
$$=500$$
Equation of the circle is
$$ (x-26)^2+(y+4)^2=500 $$
To find intersection with x-axis, put
$$y=0$$
$$ (x-26)^2+16=500 $$
$$ (x-26)^2=484 $$
$$x=26\pm22$$
Thus the two intersections are $$x=4,\ 48$$
Since $$S=(4,0)$$ one intersection is the focus itself.
Hence
$$Q=(48,0)$$
Now coordinates are
$$P=(16,16)$$
$$S=(4,0)$$
$$Q=(48,0)$$
The base
$$SQ=48-4=44$$
and the perpendicular height from $$P$$ to the x-axis is $$16$$
Therefore,
$$\text{Area of }\triangle PQS=\frac12\times44\times16$$
$$=352$$
Hence
$$\frac{\text{Area of }\triangle PQS}{11}=\frac{352}{11}$$
$$=32$$
Final Answer :
$$32$$
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