(32)

Equation of tangent to parabola $y^2=4x$ is given by  

     $y=mx+\frac{1}{m}$                                ...(i)

and equation of tangent to the circle ${(x-4)}^2+y^2=16$ is given by  

      $y=m(x-4)\pm 4\sqrt{1+m^2}$              ...(ii)

From (i) and (ii), we get  

     $\frac{1}{m}=-4m\pm 4\sqrt{1+m^2}$

$\Rightarrow 1+4m^2=\pm 4m\sqrt{1+m^2}\Rightarrow 1+16m^4+8m^2=16m^2+16m^4$

$\Rightarrow 8m^2=1\Rightarrow m=\pm \frac{1}{2\sqrt{2}}$

Point of contact of parabola is $(\frac{1}{{\left(\frac{1}{2\sqrt{2}}\right)}^2},\frac{2}{\frac{1}{2\sqrt{2}}})$                      $[\because \text{ Point of contact is} (\frac{a}{m^2},\frac{2a}{m}\left)\right]$

i.e., $(8, 4\sqrt{2})$

Now, length of tangent $PQ=\sqrt{{(8-4)}^2+{\left(4\sqrt{2}\right)}^2-16}$$=\sqrt{32}$

$\therefore$   $P{Q}^2=32$

(16)

Given, the $x$-intercept of a focal chord of the parabola $y^2=8x+4y+4$ is 3.

$y^2=8x+4y+4\Rightarrow {(y-2)}^2=8(x+1)\Rightarrow Y^2=4·2·X$

$Y=y-2, X=x+1, a=2$

For focus $x+1=2$ and $y-2=0\Rightarrow x=1$ and $y=2$

$\therefore$ $\text{Focus is }(1,2)$

Equation of line passing through $(1,2)$ is  

$y-2=m(x-1),$ $(3,0)$ lies on the above line,  

$\therefore$  $-2=m(3-1)\Rightarrow m=-1$

$\therefore$  Equation of line is $y-2=-1(x-1)$

i.e., $x+y-3=0$   $\therefore$ $y=3-x$

Now put the value of $y$ in equation of parabola ${(3-x)}^2=8x+4(3-x)+4$

$\Rightarrow 9+x^2-6x=8x+12-4x+4$  $\therefore$  $x=5+4\sqrt{2}, 5-4\sqrt{2}$

So, $y=-2-4\sqrt{2},$  $4\sqrt{2}-2$

So, length of focal chord is  

$=\sqrt{{(4\sqrt{2}+4\sqrt{2})}^2+{(-2-4\sqrt{2}-4\sqrt{2}+2)}^2}$

$=\sqrt{128+128}=\sqrt{256}=16$

(146)

As $P(b,c)$ lies on parabola  

So,  $c^2=2ab$                                           ...(i)

Now, equation of tangent to parabola $y^2=2ax$ in point form is $y{y}_1=2a\left(\frac{{\displaystyle x+x_1}}{{\displaystyle 2}}\right), \text{where }(x_1,y_1)\equiv (b,c)$

$\Rightarrow yc=a(x+b)$

For point $B$, put $y=0$, we get $x=-b$

So, area of $∆PBA$, $\frac{1}{2}\times AB\times AP=16 \text{(Given)}$

$\Rightarrow \frac{1}{2}\times 2b\times c=16\Rightarrow bc=16$

As $b$ and $c$ are natural numbers, possible values of $(b,c)$ are $(1,16),(2,8),(4,4),(8,2),(16,1)$

Now from equation (i) $a=\frac{c^2}{2b}$ and $a\in N,$ So value of $(b,c)$ are $(1,16), (2,8)$ and $(4,4)$. Now, values of $a$ are $128,16$ and $2.$

Hence, sum of values of $a$ is $146$.