Question

The slope of the tangent to the curve $y=x^{3}+x+54$ which also passes through the origin is

Answer 1

Let point $P\left(x_{1}, y_{1}\right)$ be the point on this curve such that the tangent of $P$ passes through the origin.
Equation of tangent at $P$ is $y-y_{1}=\left(3 x_{1}^{2}+1\right)\left(x-x_{1}\right)$ It passes through (0,0) so $y_{1}=\left(3 x_{1}^{2}+1\right) x_{1}$ $\Rightarrow x_{1}^{3}+x_{1}+54=3 x_{1}^{3}+x_{1}$
$\Rightarrow x_{1}^{3}=27 \Rightarrow x_{1}=3$
So, $\left.\frac{d y}{d x}\right|_{\left(x_{1}, y_{1}\right)}=3 x_{1}^{2}+1=3(9)+1=28$