The equation of the tangent to the curve y = x + 4/x², that is parallel to the x-axis, is (A) y = 1

Q: Wait, why do we set the derivative to zero? Can't we just look at the graph?

While graphing is a fantastic tool for intuition, JEE problems demand analytical precision. The derivative $\frac{dy}{dx}$ represents the instantaneous slope of the curve. A line parallel to the x-axis is horizontal, meaning it has a slope of zero. Therefore, the only way to find the exact point where the tangent is horizontal is to solve for where the derivative is zero.

Q: I got $x^3 = 8$, but what if I had $x^3 = -8$? Would the answer change?

That is an excellent observation! If the equation had resulted in $x^3 = -8$, then $x$ would be $-2$. You would then substitute $x = -2$ into the original function to find the corresponding $y$-coordinate. Always remember to solve for $x$ first, and then use the original function to find the $y$-coordinate of the point of tangency.

Q: Why do we plug $x=2$ back into the original equation and not the derivative?

The derivative tells us the slope at a point, but the original equation $y = x + \frac{4}{x^2}$ tells us the actual position (the coordinates) of the point on the curve. Since the tangent line must pass through the point on the curve, we need the $y$-coordinate from the original function to define the line's equation.

Question

The equation of the tangent to the curve $y = x + \frac{4}{x^2}$, that is parallel to the $x$-axis, is

Options:

(A) $y = 1$

(B) $y = 2$

(C) $y = 3$

(D) $y = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a landscape defined by the curve $y = x + \frac{4}{x^2}$. This curve is not just a collection of points; it is a path.

Our mission is to find a specific tangent line—a line that perfectly grazes the curve and remains parallel to the x-axis.

Because a line parallel to the x-axis is a horizontal line, it must have a slope of zero. This is our anchor for the entire calculation.

### The Calculus Toolkit: Differentiation

To find the slope of our curve at any point, we turn to the derivative. The derivative $\frac{dy}{dx}$ acts as our mathematical microscope, revealing the steepness of the curve at any given moment.

Before we differentiate, let us rewrite the function to simplify the process:
$$y = x + 4x^{-2}$$

Applying the power rule, the derivative becomes:
$$\frac{dy}{dx} = 1 - 8x^{-3}$$

This expression, which can be written as $1 - \frac{8}{x^3}$, represents the slope at any point $x$.

### The Algebraic Bridge: Solving for the Point of Contact

We established that for the tangent to be parallel to the x-axis, the slope must be zero. We set our derivative to zero:
$$1 - \frac{8}{x^3} = 0$$

This leads us to the equation:
$$1 = \frac{8}{x^3}$$

Solving for $x$, we find $x^3 = 8$, which yields the x-coordinate of the point of tangency:
$$x = 2$$

### The Final Destination: Defining the Line

Now that we have $x = 2$, we need the y-coordinate to fully define our point. We substitute $x = 2$ back into our original function:
$$y = 2 + \frac{4}{2^2}$$

Calculating this, we get:
$$y = 2 + \frac{4}{4} = 2 + 1 = 3$$

The point of tangency is $(2, 3)$. Since our tangent line is horizontal and passes through $(2, 3)$, its equation must be **$y = 3$**.

This line perfectly touches the curve at $(2, 3)$ and never deviates from its horizontal path. You have successfully navigated the relationship between derivatives, slopes, and geometry.

Animated Solution for Mathematics - Differentiation: The equation of the tangent to the curve $y = x + \frac{4}{x ^{2}}$ , that is parallel to the $x$ -axis, is

Select Answer:

Visualized Solution

Visualizing the Problem

Condition for Parallel Tangent

Slope of a Curve

Preparing for Differentiation

Differentiating the Function

Applying the Power Rule

Simplifying the Derivative

Setting Derivative to Zero

Solving for $x$

Finding the x-coordinate

Finding the y-coordinate

Calculating the y-coordinate

Equation of the Tangent Line

Final Answer

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Calculus Toolkit

The Algebraic Bridge

The Final Destination

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of the tangent to the curve y=x+x24​, that is parallel to the x-axis, is