A vector of magnitude 10 along the normal to the curve $3{\mathrm{x}}^2+8\mathrm{xy}+2{\mathrm{y}}^2-3=0$ at its point P(1,0) can

Differentiate the curve 

$\begin{array}{l}6\mathrm{x}+8\left({\mathrm{xy}}_1+\mathrm{y}\right)+4{\mathrm{yy}}_1=0\\ {\mathrm{m}}_{\mathrm{T}}\text{ at }(1,0)\text{ is }6+8\left({\mathrm{y}}_1\left(0\right)\right)=0\\ {\mathrm{y}}_1\left(0\right)=-\frac{3}{4}\\ {\mathrm{m}}_{\mathrm{N}}=\frac{4}{3}\\ \text{ Unit vector }=\pm \frac{(3\widehat{\mathrm{i}}+4\widehat{\mathrm{j}})}{5}\end{array}$

$\text{ Again normal vector of magnitude }10=\pm (6\widehat{i}+8\widehat{j})$