Question

If the normal to the curve $x=t-1,y=3t^2-6$ at the point $(1,6)$ makes intercepts $a$ and $b$ on $x$ and $y-$axis respectively, then the value $a+12b$ is

Answer 1

Given point is corresponding to $t=2$ and $\frac{dy}{dx}=6t$
$\Rightarrow$ Slope of normal at $t=2$ is $-\frac{1}{12}$
$\therefore$ Equation of normal is $y-6=-\frac{1}{12}(x-1).$
$x-$ intercept is $a=73$
$y-$intercept is $b=\frac{73}{12}$
$\Rightarrow a+12b=146$