the tangent at the point (x₁, y₁) on the curve y = x³ + 3x² + 5 passes through the origin, then (x₁,

Q: Wait, why do we equate the two expressions for $y_1$?

That is a great question! We have two constraints on the point $(x_1, y_1)$. First, it must lie on the curve, which gives us $y_1 = x_1^3 + 3x_1^2 + 5$. Second, the tangent at this point must pass through the origin, which leads to the condition $y_1 = x_1 \cdot f'(x_1)$. Since both expressions must hold true simultaneously for the point to satisfy both conditions, equating them allows us to find the specific $x_1$ that makes this geometry possible.

Q: How do I know if there are other points on the curve where the tangent passes through the origin?

Excellent curiosity! By equating the two expressions, we derived the cubic equation $2x_1^3 + 3x_1^2 - 5 = 0$. Solving this cubic reveals the roots. We found $x_1 = 1$ by inspection. Factoring out $(x_1 - 1)$ leaves us with a quadratic $2x_1^2 + 5x_1 + 5 = 0$. If you calculate the discriminant ($D = b^2 - 4ac$), you will see it is $25 - 40 = -15$, which is less than zero. This confirms there are no other real roots, meaning $(1, 9)$ is the unique point.

Q: Why does the question ask which curve it does NOT lie on?

This is a classic JEE-style 'negative' question designed to test your verification skills. Instead of just finding the point, you must apply the point to each option. It forces you to be precise with your arithmetic. If you find the point $(1, 9)$ and it satisfies three of the options, the one that fails is your answer. It is a way to ensure you haven't just guessed the answer!

Question

the tangent at the point $(x_1, y_1)$ on the curve $y = x^3 + 3x^2 + 5$ passes through the origin, then $(x_1, y_1)$ does NOT lie on the curve :

Options:

(A) $x^2 + \frac{y^2}{81} = 2$

(B) $\frac{y^2}{9} - x^2 = 8$

(C) $y = 4x^2 + 5$

(D) $\frac{x}{3} - y^2 = 2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a smooth, winding path defined by the function $y = x^3 + 3x^2 + 5$. You are looking for a specific location on this path—a point $(x_1, y_1)$ where the tangent line passes perfectly through the origin $(0, 0)$.

This is a quest to find a specific alignment between a curve and the coordinate system.

### The Calculus Toolkit

To find this point, we first determine the slope of our path at any given location using the derivative. By differentiating $y = x^3 + 3x^2 + 5$ with respect to $x$, we obtain:

$$ \frac{dy}{dx} = 3x^2 + 6x $$

This expression defines the slope $m$ at any point $x$. Specifically, at our mystery point $(x_1, y_1)$, the slope is $m = 3x_1^2 + 6x_1$.

### The Origin Constraint

We construct the tangent line using the point-slope form $y - y_1 = m(x - x_1)$. Substituting our slope $m$, we get:

$$ y - y_1 = (3x_1^2 + 6x_1)(x - x_1) $$

Because this line passes through the origin $(0, 0)$, we substitute $x = 0$ and $y = 0$ into the equation:

$$ 0 - y_1 = (3x_1^2 + 6x_1)(0 - x_1) $$

Simplifying this expression, we find $-y_1 = -3x_1^3 - 6x_1^2$, which rearranges to:

$$ y_1 = 3x_1^3 + 6x_1^2 $$

### The Algebraic Bridge

We now have two expressions for $y_1$. From the original curve, $y_1 = x_1^3 + 3x_1^2 + 5$, and from the tangent condition, $y_1 = 3x_1^3 + 6x_1^2$. Setting them equal, we obtain:

$$ x_1^3 + 3x_1^2 + 5 = 3x_1^3 + 6x_1^2 $$

Bringing all terms to one side yields the cubic equation:

$$ 2x_1^3 + 3x_1^2 - 5 = 0 $$

By testing simple integers, we see that $x_1 = 1$ satisfies the equation ($2 + 3 - 5 = 0$). Factoring the cubic confirms this is the only real solution.

Plugging $x_1 = 1$ back into the curve equation, we find $y_1 = 1^3 + 3(1)^2 + 5 = 9$. Thus, our mystery point is **$(1, 9)$**.

### The Final Verification

We test this point against the provided conditions. For the first, $1^2 + \frac{9^2}{81} = 1 + 1 = 2$ (True). For the second, $\frac{9^2}{9} - 1^2 = 9 - 1 = 8$ (True).

For the third, $9 = 4(1)^2 + 5 = 9$ (True). However, for the fourth, $\frac{1}{3} - 9^2 = \frac{1}{3} - 81$, which is not $2$.

We have successfully identified the point $(1, 9)$ and verified the geometric constraints.

Animated Solution for Mathematics - Differentiation: the tangent at the point $(x_{1}, y_{1})$ on the curve $y = x^{3} + 3 x^{2} + 5$ passes through the origin, then $(x_{1}, y_{1})$ does NOT lie on the curve :

Select Answer:

Visualized Solution

Identify the Curve and Point

Point Satisfies the Curve

Slope of the Tangent

Equation of the Tangent

Apply the Origin Condition

Simplify the Origin Condition

Equate the Two Equations

Form the Cubic Equation

Solve the Cubic Equation

Calculate the y-coordinate

Verify the Options

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Calculus Toolkit

The Origin Constraint

The Algebraic Bridge

The Final Verification

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