Point (x1, y1) lies on the curve
\(y = x^3 + 3x^2 + 5\)
\(\therefore y_1 = {x_1}^3 + 3{x_1}^2 + 5\) ....(1)
\(\frac{dy}{dx} = 3x^2 + 6x\)
\(\therefore \left(\frac{dy}{dx}\right) = 3x^2 + 6x\)
Tangent is passes through origin.
∴ Equation of tangent is
\(y - 0 = \left(\frac{dy}{dx}\right) _{(x_1, y_1)} (x - 0)\)
⇒ \(y = (3x^2 + 6x )(x - 0)\)
⇒ \(y = 3x^3 + 6x^2\) .....(2)
Point (x1, y1) lies on curve (2).
\(\therefore y_1 = 3{x_1}^3 + 6{x_1}^2\)
\(= 3({x_1}^3 + 2{x_1}^2) \)
\(=3({y_1} - {x_1}^2 - 5) \)
\(= 3y_1 - 3{x_1}^2 - 15\)
⇒ \(3{x_1}^2 = 2y_1 - 15\)
Hence, (x1, y1) lies on 3x2 = 2y - 15.