Let the normals at all the points on a given curve pass through a fixed point $(a, b)$. If the curve passes through $(3,-3)$ and $(4,-2 \sqrt{2})$, and given that $a-2 \sqrt{2} \mathrm{~b}=3$, then $\left(\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{ab}\right)$ is equal to
Let the equation of normal is $Y-y=-\frac{1}{m}(X-x)$, where, $m=\frac{d y}{d x}$ As it passes through $(a, b)$
$b-y=-\frac{1}{m}(a-x)=-\frac{d x}{d y}(a-x)$
$\Rightarrow(b-y) d y=(x-a) d x$
by $-\frac{y^{2}}{2}=\frac{z^{2}}{2}-a x+c$
It passes through $(3,-3) \&(4,-2 \sqrt{2})$
$\therefore \quad-3 b-\frac{9}{2}=\frac{9}{2}-3 a+c$
$\Rightarrow-6 b-9=9-6 a+2 c$
$\Rightarrow 6 a-6 b-2 c=18$
$\Rightarrow 3 a-3 b-c=9$
Also $-2 \sqrt{2 b}-4=8-4 a+c$
$4 a-2 \sqrt{2} b-c=12$
Also $a-2 \sqrt{2} b=3 \quad \ldots$ (iv) (given)
$4 a-2 \sqrt{2} b-c=12$
(ii) $-($ iii $) \Rightarrow-a+(2 \sqrt{2}-3) b=-3 \quad \ldots(v)$
$(\mathrm{iV})+(\mathrm{v}) \Rightarrow \mathrm{b}=0, \quad \mathrm{a}=3$
$\therefore a^{2}+b^{2}+a b=9$