If the tangent to the curve y = x/(x²-3), x ∈ R, x ≠ ±√3, at a point (α, β) ≠ (0, 0) on it is parall

Q: Wait, why do we reject the solution $\alpha = 0$?

Great catch! The problem explicitly states that the point $(\alpha, \beta) \neq (0, 0)$. If we had $\alpha = 0$, the point would be $(0, 0)$, which is the origin. While mathematically $\alpha = 0$ is a root of the polynomial, it violates the problem's domain constraint. Always keep an eye on those 'fine print' conditions!

Q: Why can't we just use energy conservation or some other physics principle?

I love the enthusiasm! However, this is a pure coordinate geometry and calculus problem. We aren't dealing with forces or energy here, but rather the local behavior of a function. The 'energy' here is the slope, and the 'force' is the derivative! We must stick to the language of calculus to solve this.

Q: Is there a geometric interpretation of the absolute value $|6\alpha + 2\beta| = 19$?

Think of it as a measure of distance or a specific linear combination of coordinates. In JEE problems, these expressions often appear to test if you can manipulate the coordinates of the points you've found. It's not necessarily a distance formula, but rather a property of the specific points lying on that curve.

Question

If the tangent to the curve $y = \frac{x}{x^2-3}, x \in R, x 
e \pm\sqrt{3}$, at a point $(\alpha, \beta) 
e (0, 0)$ on it is parallel to the line $2x + 6y - 11 = 0$, then :

Options:

(A) |6\alpha + 2\beta| = 19

(B) |2\alpha + 6\beta| = 11

(C) |6\alpha + 2\beta| = 9

(D) |2\alpha + 6\beta| = 19

Sigma Percentile · Accepted Answer

### The Dance of the Tangent: A Journey into Calculus

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are embarking on a journey to understand the intimate relationship between a curve and a line.

Imagine you are standing on a roller coaster track defined by the function $y = \frac{x}{x^2-3}$. You are looking for a specific spot, a point $(\alpha, \beta)$, where if you were to jump off, your trajectory would be perfectly parallel to a nearby path defined by the line $2x + 6y - 11 = 0$.

This is the essence of the problem: finding the point of perfect alignment.

### Phase 1: The Geometry of Parallelism

Before we dive into the heavy algebra, let us ground ourselves in geometry. What does it mean for two lines to be parallel? It means they share the same inclination.

In the world of coordinate geometry, this is synonymous with having the same slope. Our reference line is $2x + 6y - 11 = 0$.

To find its slope, we perform a simple transformation into the slope-intercept form, $y = mx + c$. Rearranging our equation, we get $6y = -2x + 11$, which simplifies to:

$$y = -\frac{1}{3}x + \frac{11}{6}$$

The slope, $m_L$, is clearly $-\frac{1}{3}$. This is our target; we need the tangent to our curve to have this exact slope.

### Phase 2: The Calculus Engine

Now, we turn to the curve. How do we find the slope of a tangent at any point on $y = \frac{x}{x^2-3}$? We use the derivative, $\frac{dy}{dx}$.

Because our function is a ratio of two polynomials, we must summon the Quotient Rule. Recall the mantra: "Low d-High minus High d-Low, over the square of what's below."

Applying this, we set $u = x$ and $v = x^2 - 3$. The derivative of $u$ is $1$, and the derivative of $v$ is $2x$. Plugging these into the formula $\frac{v u' - u v'}{v^2}$, we get:

$$\frac{dy}{dx} = \frac{(x^2-3)(1) - x(2x)}{(x^2-3)^2}$$

Simplifying the numerator, we find $x^2 - 3 - 2x^2$, which leaves us with $-x^2 - 3$. Thus, the slope of our tangent at any point $x$ is:

$$\frac{-(x^2 + 3)}{(x^2 - 3)^2}$$

### Phase 3: The Algebraic Dance

We are now at the heart of the problem. We equate our derivative to the slope of the line:

$$\frac{-(x^2 + 3)}{(x^2 - 3)^2} = -\frac{1}{3}$$

Notice the negative signs on both sides? They cancel out, leaving us with $\frac{x^2 + 3}{(x^2 - 3)^2} = \frac{1}{3}$. Now, we cross-multiply to obtain:

$$3(x^2 + 3) = (x^2 - 3)^2$$

Expanding both sides, we get $3x^2 + 9 = x^4 - 6x^2 + 9$. The constant $9$ vanishes from both sides, a beautiful simplification!

We are left with $x^4 - 9x^2 = 0$. Factoring this, we get $x^2(x^2 - 9) = 0$. This yields $x = 0$ or $x^2 = 9$. Since the problem implies a valid tangent, we discard $x = 0$ and accept $x = \pm 3$.

### Phase 4: The Final Verification

We have found our $\alpha$ values: $3$ and $-3$. Now, we find the corresponding $\beta$ values.

For $\alpha = 3$, $\beta = \frac{3}{3^2 - 3} = \frac{3}{6} = \frac{1}{2}$. For $\alpha = -3$, $\beta = \frac{-3}{(-3)^2 - 3} = -\frac{3}{6} = -\frac{1}{2}$.

We have two points: $(3, \frac{1}{2})$ and $(-3, -\frac{1}{2})$. Finally, we test the expression $|6\alpha + 2\beta|$.

For the first point, $|6(3) + 2(\frac{1}{2})| = |18 + 1| = 19$. For the second point, $|6(-3) + 2(-\frac{1}{2})| = |-18 - 1| = |-19| = 19$.

The result is consistent. You have successfully navigated the curve and found the point of alignment. The final answer is **19**.

Animated Solution for Mathematics - Differentiation: If the tangent to the curve $y = \frac{x}{x ^{2} - 3}, x \in R, x \neq = \pm 3$ , at a point $(α, β) \neq = (0, 0)$ on it is parallel to the line $2 x + 6 y - 11 = 0$ , then :

Select Answer:

Visualized Solution

Visualizing the Curve and Tangent

The Concept of Parallel Slopes

Differentiating the Curve

Applying the Quotient Rule

Simplifying the Derivative

Slope of the Reference Line

Equating the Slopes

Cross-Multiplying

Expanding the Equation

Finding the Values of $α$

Calculating the Corresponding $β$ Values

Verifying the Given Options

The Sigma Insight: Tangents, Normals and Rate Measure

The Dance of the Tangent

Phase 1

Phase 2

Phase 3

Phase 4

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