Question:
Equation of tangent to the curve
\[
y=x+\frac{4}{x^2}
\]
which is parallel to \(x\)-axis is
Equation of tangent to the curve
\[
y=x+\frac{4}{x^2}
\]
which is parallel to \(x\)-axis is
Show Hint
A tangent parallel to the \(x\)-axis has slope \(0\). Hence, put \(\frac{dy}{dx}=0\) and then find the corresponding point on the curve.
Updated On: Jun 26, 2026
- \(y=8\)
- \(y=0\)
- \(y=3\)
- \(y=2\)
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The Correct Option is C
Solution and Explanation
Step 1: Differentiate the curve.
Given, \[ y=x+\frac{4}{x^2} \] \[ y=x+4x^{-2} \] Differentiating with respect to \(x\), \[ \frac{dy}{dx}=1-8x^{-3} \] \[ \frac{dy}{dx}=1-\frac{8}{x^3}. \]
Step 2: Use the condition for tangent parallel to \(x\)-axis.
A tangent parallel to the \(x\)-axis has slope \(0\).
So, \[ \frac{dy}{dx}=0. \] Therefore, \[ 1-\frac{8}{x^3}=0 \] \[ \frac{8}{x^3}=1 \] \[ x^3=8 \] \[ x=2. \]
Step 3: Find the corresponding value of \(y\).
Substitute \(x=2\) in \[ y=x+\frac{4}{x^2}. \] \[ y=2+\frac{4}{2^2} \] \[ y=2+\frac{4}{4} \] \[ y=2+1 \] \[ y=3. \]
Step 4: Write the tangent equation.
Since the tangent is parallel to the \(x\)-axis, its equation is of the form \[ y=c. \] Here, \[ c=3. \] So, the tangent is \[ y=3. \]
Step 5: Final conclusion.
Therefore, \[ \boxed{y=3} \]
Given, \[ y=x+\frac{4}{x^2} \] \[ y=x+4x^{-2} \] Differentiating with respect to \(x\), \[ \frac{dy}{dx}=1-8x^{-3} \] \[ \frac{dy}{dx}=1-\frac{8}{x^3}. \]
Step 2: Use the condition for tangent parallel to \(x\)-axis.
A tangent parallel to the \(x\)-axis has slope \(0\).
So, \[ \frac{dy}{dx}=0. \] Therefore, \[ 1-\frac{8}{x^3}=0 \] \[ \frac{8}{x^3}=1 \] \[ x^3=8 \] \[ x=2. \]
Step 3: Find the corresponding value of \(y\).
Substitute \(x=2\) in \[ y=x+\frac{4}{x^2}. \] \[ y=2+\frac{4}{2^2} \] \[ y=2+\frac{4}{4} \] \[ y=2+1 \] \[ y=3. \]
Step 4: Write the tangent equation.
Since the tangent is parallel to the \(x\)-axis, its equation is of the form \[ y=c. \] Here, \[ c=3. \] So, the tangent is \[ y=3. \]
Step 5: Final conclusion.
Therefore, \[ \boxed{y=3} \]
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