Applications Of Derivatives

Step 1: Differentiate the curve.
Given: Differentiate with respect to :

Step 2: Find slope of tangent at .
So, slope of tangent at is .

Step 3: Find slope of normal.
The slope of the normal is the negative reciprocal of the slope of the tangent.

Step 4: Match with options.
The correct answer is option (C) .

Step 1: Implicit Differentiation. Differentiate both sides of the equation with respect to : This gives: Solve for :

Step 2: Condition for the normal being parallel to the x-axis. The normal is parallel to the x-axis if the slope of the normal is 0. The slope of the normal is the negative reciprocal of the slope of the tangent. Therefore, we need: This implies: Thus, .

Step 3: Find the corresponding value of . Substitute into the original equation: Conclusion: The points on the curve where the normal is parallel to the x-axis are and .

Step 1: Slope of the given line. Equation: . So, slope = .

Step 2: Differentiate the curve. Curve: Differentiate both sides w.r.t. :

Step 3: Condition for parallelism. We need .

Step 4: Corresponding points. If . Also, from curve . So, , . For , possible .

Step 5: Equation of tangent. Equation of tangent: , with . At : At :

Final Answer:

Step 1: Slope of given tangent.
Equation of tangent line:

Step 2: Differentiate the curve.
Curve:

Step 3: Equating slope.
At tangent point, slope :

Step 4: Find corresponding values.
For : For :

Step 5: Check if points lie on tangent line.
Tangent: . For : RHS = , ✅ lies on line. For : RHS = , ❌ not valid.

Final Answer: