(72)

We have, $L : \sqrt{2}x+y=\alpha$

 $x^2+y^2=3$         ... (i)

  $x^2=2y$             ... (ii)

Equation (i) and (ii) intersect at $(\sqrt{2}, 1)$ in first quadrant.

$\therefore P=(\sqrt{2},1) \mathrm{and} \alpha =\sqrt{2}\times \sqrt{2}+1=3$

$\Rightarrow \alpha =3$

Radius of Circle $C_1$ and $C_2=2\sqrt{3}$

We centre $C_1$ & $C_2$ of two circle are $(0, y_1)$ and $(0, y_2)$ respectively.

We know that length of perpendicular from centre to the tangent = Radius of circle

$\therefore$$\left|\frac{\sqrt{2}\times 0+y–3}{\sqrt{3}}\right|$$=2\sqrt{3}$

$\Rightarrow |y–3|=6 \Rightarrow y=9, –3$

$\therefore y_1=–3, y_2=9$

$\therefore$  Centre of $C_1$ and $C_2$ are $Q_1(0, –3)$ and $Q_2(0, 9)$ respectively.

$Q_1{Q}_2=12$ units, Height of triangle $=\sqrt{2}$ units

A = Area of triangle $P{Q}_1{Q}_2=\frac{1}{2}\times 12\times \sqrt{2}$

                                   $[\because \mathrm{Area} \mathrm{of} △P{Q}_1{Q}_2=\frac{1}{2}\times Q_1{Q}_2\times \mathrm{height}]$

$=6\sqrt{2} \mathrm{sq}. \mathrm{units}$

$\Rightarrow A^2=72$.

(192)

We $\alpha = t^2, \beta = 2t$

Now, equation of chord of the parabola $x^2=8y$, bisected at $(1, \frac{5}{4})$ is $x·1–2\times 2(y+\frac{5}{4})=1–10$

$\Rightarrow x–4y–5=–9 \Rightarrow x–4y+4=0$

Now, $(\alpha , \beta )$ satisfies the above equation

$\therefore t^2–8t+4=0$            ... (i)

Now, $(\alpha –28)(\beta –8)=(t^2–28)(2t–8)$

$=(8t–4–28)(2t–8)$           (Using (i))

$=16t^2–64t–64t+256=16(t^2–8t+16)$

$=16(t^2–8t+4+12)=192$.