Let the normal at a P on the curve y² - 3x² + y + 10 = 0 intersect the y-axis at (0, 3/2). If m is t

Q: Wait, why can we just cancel $x_1$ in the equation $m \times m_n = -1$?

That is a great observation! We can cancel $x_1$ because if $x_1$ were zero, the normal would be a vertical line (undefined slope), but the problem states the normal passes through $(0, 1.5)$, which would imply the normal is the y-axis itself. If the normal were the y-axis, the tangent at $P$ would be horizontal, meaning $m=0$. However, our slope formula $m = \frac{6x_1}{2y_1 + 1}$ would then give $m=0$ only if $x_1=0$. Checking the curve equation for $x_1=0$, we get $y^2 + y + 10 = 0$, which has no real roots. Thus, $x_1$ cannot be zero, and it is perfectly safe to cancel it.

Q: Why is the normal slope $\frac{y_1 - 1.5}{x_1}$?

The normal is a straight line passing through two points: the point of contact $P(x_1, y_1)$ and the given intersection point $A(0, 1.5)$. The slope of any line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\frac{y_2 - y_1}{x_2 - x_1}$. Substituting our points, we get $\frac{1.5 - y_1}{0 - x_1}$, which simplifies to $\frac{y_1 - 1.5}{x_1}$. It is just the standard two-point slope formula!

Q: Why do we need the absolute value $|m|$ at the end?

The problem asks for $|m|$ because the curve is symmetric with respect to the y-axis. When we solved for $x_1$, we found $x_1 = \pm 2$. This means there are two points on the curve that satisfy the condition—one with a positive x-coordinate and one with a negative x-coordinate. These two points result in tangent slopes that are equal in magnitude but opposite in sign ($4$ and $-4$). The absolute value ensures we capture the magnitude of the slope regardless of which point $P$ we are considering.

Question

Let the normal at a $P$ on the curve $y^2 - 3x^2 + y + 10 = 0$ intersect the y-axis at $(0, \frac{3}{2})$. If $m$ is the slope of the tangent at $P$ to the curve, then $|m|$ is equal to____________.

Sigma Percentile · Accepted Answer

### The Geometry of the Tangent and the Normal

Welcome, fellow explorer of mathematics. Today, we are going to dissect a beautiful problem that sits at the intersection of calculus and coordinate geometry.

We are given a curve defined by the implicit equation $y^2 - 3x^2 + y + 10 = 0$. Our mission is to find the slope of the tangent at a specific point $P$ on this curve, given that the normal at $P$ passes through a specific point on the y-axis.

Let us embark on this journey step by step.

### Phase 1: The Power of Implicit Differentiation

First, we must understand the nature of our curve. It is an implicit relation, which means $y$ is not explicitly defined as a function of $x$.

To find the slope of the tangent, we need the derivative $\frac{dy}{dx}$. We apply implicit differentiation to the equation $y^2 - 3x^2 + y + 10 = 0$.

Differentiating term by term with respect to $x$, we get:

$$\frac{d}{dx}(y^2) - \frac{d}{dx}(3x^2) + \frac{d}{dx}(y) + \frac{d}{dx}(10) = 0$$

Using the chain rule, this becomes $2y \frac{dy}{dx} - 6x + \frac{dy}{dx} = 0$. By grouping the terms containing $\frac{dy}{dx}$, we find that $\frac{dy}{dx}(2y + 1) = 6x$.

Thus, the slope of the tangent at any point $(x_1, y_1)$ on the curve is given by:

$$m = \frac{6x_1}{2y_1 + 1}$$

This is our primary tool for understanding the local behavior of the curve at point $P$.

### Phase 2: The Normal's Secret

Now, let us turn our attention to the normal line. We are told that the normal at $P(x_1, y_1)$ intersects the y-axis at $(0, 1.5)$.

A normal line is, by definition, perpendicular to the tangent line at the point of contact. The slope of this normal line, $m_n$, can be calculated using the two-point slope formula:

$$m_n = \frac{y_1 - 1.5}{x_1 - 0} = \frac{y_1 - 1.5}{x_1}$$

This simple geometric relationship is the key that unlocks the entire problem.

### Phase 3: The Orthogonality Condition

We know that for two lines to be perpendicular, the product of their slopes must be $-1$. Therefore, $m \cdot m_n = -1$.

Substituting our expressions, we have:

$$\left( \frac{6x_1}{2y_1 + 1} \right) \cdot \left( \frac{y_1 - 1.5}{x_1} \right) = -1$$

This is the moment of truth. Notice how the $x_1$ terms in the numerator and denominator cancel out beautifully, leaving us with a much simpler equation:

$$\frac{6(y_1 - 1.5)}{2y_1 + 1} = -1$$

Multiplying both sides by the denominator, we get $6y_1 - 9 = -(2y_1 + 1)$, which simplifies to $6y_1 - 9 = -2y_1 - 1$. Solving for $y_1$, we find $8y_1 = 8$, or simply $y_1 = 1$.

### Phase 4: The Final Convergence

With $y_1 = 1$ in hand, we return to the original curve equation to find the corresponding $x_1$. Substituting $y_1 = 1$ into $y^2 - 3x^2 + y + 10 = 0$, we get:

$$1^2 - 3x_1^2 + 1 + 10 = 0$$

This simplifies to $12 - 3x_1^2 = 0$. This leads to $3x_1^2 = 12$, so $x_1^2 = 4$, meaning $x_1 = \pm 2$.

Finally, we calculate the slope $m$ using our earlier formula $m = \frac{6x_1}{2y_1 + 1}$:

For $x_1 = 2$, $m = \frac{6(2)}{2(1) + 1} = \frac{12}{3} = 4$. 
For $x_1 = -2$, $m = \frac{6(-2)}{3} = -4$.

In both cases, the **absolute value $|m|$ is 4**. We have successfully navigated the complexity of the curve to arrive at a clean, elegant result.

Animated Solution for Mathematics - Differentiation: Let the normal at a $P$ on the curve $y^{2} - 3 x^{2} + y + 10 = 0$ intersect the y-axis at $(0, \frac{3}{2})$ . If $m$ is the slope of the tangent at $P$ to the curve, then $∣ m ∣$ is equal to____________.

Enter Numerical Value:

Visualized Solution

Define Point $P$ and the Curve

Implicit Differentiation

Calculate Tangent Slope $m$

Geometric Slope of Normal

Orthogonality Condition

Solving for $y_{1}$

Finding $x_{1}$

Final Calculation of $∣ m ∣$

The Sigma Insight: Tangents, Normals and Rate Measure

The Geometry of the Tangent and the Normal

Phase 1

Phase 2

Phase 3

Phase 4

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Enter Numerical Value:

Visualized Solution

Define Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and the Curve

Implicit Differentiation

Calculate Tangent Slope .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

Geometric Slope of Normal

Orthogonality Condition

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y1​

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x1​

Final Calculation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣m∣