Question

The co-ordinates of the point on the curve \(y = xlogx\) at which the normal is parallel to the line \(2x-2y = 3\) are...

• A. \((0,0)\)
• B. \((e,e)\)
• C. \((e^2,2e^2)\)
• D. \((e^{-2},-2e^{-2})\)

Hint:

The normal is parallel to slope 1, so the tangent has slope -1.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The normal at a point is perpendicular to the tangent. The line \(2x-2y=3\) has slope \(1\), so the normal must have slope \(1\), which means the tangent has slope \(-1\).

Step 2: Slope of the curve:
\(y=x\ln x\), so \(\dfrac{dy}{dx}=\ln x+1\).

Step 3: Condition:
Tangent slope \(=-1\): \(\ln x+1=-1\), so \(\ln x=-2\) and \(x=e^{-2}\).

Step 4: Find y:
\(y=x\ln x=e^{-2}\cdot(-2)=-2e^{-2}\).

Step 5: Choose:
The point is \((e^{-2},-2e^{-2})\), option (D).

Final Answer:
The point is (e^-2, -2e^-2). \[ \boxed{(e^{-2},-2e^{-2})} \]