Question

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 $a-2\sqrt{2}b=3$, then $a^2+b^2+ab$ is equal to

• A. $7$
• B. $9$
• C. $6$
• D. $8$

Answer 1

The correct option is B $9$

Let the equation of normal is $Y-y=-\frac{1}{m}(X-x),\text{where}m=\frac{dy}{dx}$
As it passes through $(a,b)$
$\Rightarrow b-y=-\frac{1}{m}(a-x)=-\frac{dx}{dy}(a-x)$
$\Rightarrow (b-y)dy=(x-a)dx$
$\Rightarrow by-\frac{y^2}{2}=\frac{x^2}{2}-ax+c....\left(i\right)$
It passes through $(3,-3)\&(4,-2\sqrt{2})$
$\therefore -3b-\frac{9}{2}=\frac{9}{2}-3a+c$
$\Rightarrow 3a-3b-c=\mathrm{9...}\left(ii\right)$
Also
$-2\sqrt{2}b-4=8-4a+c$
$\Rightarrow 4a-2\sqrt{2}b-c=\mathrm{12...}\left(iii\right)$
Also $a-2\sqrt{2}b=3\dots \left(iv\right)$ (given)
From $\left(ii\right)-\left(iii\right)$
$\Rightarrow -a+(2\sqrt{2}-3)b=-3\dots \left(v\right)$
From $\left(iv\right)+\left(v\right)$
$\Rightarrow b=0,a=3$
$\therefore a^2+b^2+ab=9$