Key Concepts and Formulas

• Slope of Tangent: The slope of the tangent to a curve $y = f(x)$ at a point $(x_1, y_1)$ is given by the derivative: $m = \left(\frac{dy}{dx}\right)_{(x_1, y_1)}$.
• Equation of Tangent: The equation of the tangent line at $(x_1, y_1)$ is given by $y - y_1 = m(x - x_1)$.
• Line Through Origin: If a line passes through the origin $(0,0)$, the ratio of the y-coordinate to the x-coordinate of any point on the line gives the slope.

Step-by-Step Solution

Step 1: Identify the Given Curve and a Point on It

We are given the curve $y = x^3 + 3x^2 + 5$. Let $(x_1, y_1)$ be the point of tangency. Since $(x_1, y_1)$ lies on the curve, it must satisfy the curve's equation: $y_1 = x_1^3 + 3x_1^2 + 5 \quad \text{(Equation 1)}$

Step 2: Find the Derivative of the Curve

To find the slope of the tangent at any point $(x, y)$, we differentiate the curve's equation with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(x^3 + 3x^2 + 5)$ $\frac{dy}{dx} = 3x^2 + 6x$

Step 3: Find the Slope of the Tangent at $(x_1, y_1)$

The slope of the tangent at the point $(x_1, y_1)$ is: $m = \left(\frac{dy}{dx}\right)_{(x_1, y_1)} = 3x_1^2 + 6x_1$

Step 4: Use the Condition that the Tangent Passes Through the Origin

Since the tangent passes through the origin $(0,0)$, its slope can also be calculated as: $m = \frac{y_1 - 0}{x_1 - 0} = \frac{y_1}{x_1}$

Equating the two expressions for the slope gives us: $\frac{y_1}{x_1} = 3x_1^2 + 6x_1$ Assuming $x_1 \neq 0$ (we will check this later), we multiply by $x_1$: $y_1 = x_1(3x_1^2 + 6x_1)$ $y_1 = 3x_1^3 + 6x_1^2 \quad \text{(Equation 2)}$

Step 5: Solve the System of Equations

We now have two equations:

1. $y_1 = x_1^3 + 3x_1^2 + 5$
2. $y_1 = 3x_1^3 + 6x_1^2$

Equating the expressions for $y_1$: $x_1^3 + 3x_1^2 + 5 = 3x_1^3 + 6x_1^2$ Rearrange the terms: $2x_1^3 + 3x_1^2 - 5 = 0$

We look for integer roots. Trying $x_1 = 1$: $2(1)^3 + 3(1)^2 - 5 = 2 + 3 - 5 = 0$ So, $x_1 = 1$ is a root.

Substitute $x_1 = 1$ into Equation 1 to find $y_1$: $y_1 = (1)^3 + 3(1)^2 + 5$ $y_1 = 1 + 3 + 5 = 9$ Thus, the point of tangency is $(x_1, y_1) = (1, 9)$.

If $x_1 = 0$, then from Equation 2, $y_1 = 0$. But $(0,0)$ does not satisfy Equation 1 ($0 \neq 5$). So, $x_1 \neq 0$ is correct.

Step 6: Check Which Curve the Point $(1, 9)$ Does NOT Lie On

Substitute $x=1$ and $y=9$ into each option:

• (A) $x^2 + \frac{y^2}{81} = 2$ $(1)^2 + \frac{(9)^2}{81} = 1 + \frac{81}{81} = 1 + 1 = 2$. Satisfied.

• (B) $\frac{y^2}{9} - x^2 = 8$ $\frac{(9)^2}{9} - (1)^2 = \frac{81}{9} - 1 = 9 - 1 = 8$. Satisfied.

• (C) $y = 4x^2 + 5$ $9 = 4(1)^2 + 5$ $9 = 4 + 5 = 9$. Satisfied.

• (D) $\frac{x}{3} - y^2 = 2$ $\frac{1}{3} - (9)^2 = \frac{1}{3} - 81 = \frac{1 - 243}{3} = -\frac{242}{3} \neq 2$. Not satisfied.

Therefore, the point $(1, 9)$ does not lie on the curve given in option (D).

Common Mistakes & Tips

• Remember the point is on the curve: Don't forget that $(x_1, y_1)$ satisfies the original curve equation, leading to a system of equations.
• Careful Algebra: Double-check arithmetic, especially with cubic equations. Integer root theorem helps.

Summary

We found the point of tangency $(1, 9)$ by using the derivative to find the slope of the tangent, the condition that the tangent passes through the origin, and the fact that the point lies on the curve. Substituting this point into each of the options, we found that the point does not lie on the curve in option (A).

Final Answer

The final answer is \boxed{A}, which corresponds to option (A).