the tangent at the point (x₁, y₁) on the curve y = x³ +3x² + 5 passes through the origin, then (x₁, y₁) does NOT lie on the curve :

The tangent at (x₁, y₁) to the curve
$\begin{array}{l}\mathrm{y}={\mathrm{x}}^3+3{\mathrm{x}}^2+5\\ \mathrm{y}-{\mathrm{y}}_1=\left(3{\mathrm{x}}_1^2+6{\mathrm{x}}_1\right)\left(\mathrm{x}-{\mathrm{x}}_1\right)\text{ passing through origin }\\ -{\mathrm{y}}_1=\left(3{\mathrm{x}}_1^2+6{\mathrm{x}}_1\right)\left(-{\mathrm{x}}_1\right)\\ {\mathrm{y}}_1=\left(3{\mathrm{x}}_1^3+6{\mathrm{x}}_1^2\right)\end{array}$
And (x₁ ,y₁) lies on the curve
$\mathrm{y}={\mathrm{x}}^3+3{\mathrm{x}}^2+5$
${\mathrm{y}}_1={\mathrm{x}}_1^3+3{\mathrm{x}}_1^2+5$........(2)
From equation (1) and (2)
$2{\mathrm{y}}_1=3{\mathrm{x}}_1^2+15$
Hence the equation of curve $\mathrm{y}=\frac{3}{2}{\mathrm{x}}^2+\frac{15}{2}$

This curve does not intersect $\frac{x}{3}-y^2=2$