Question:
If the normal drawn at the point P on the curve \(y = x \log x\) is parallel to the line \(2x - 2y = 3\), then P =
If the normal drawn at the point P on the curve \(y = x \log x\) is parallel to the line \(2x - 2y = 3\), then P =
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To find the point where a normal has a given slope, equate the negative reciprocal of the derivative and solve.
Updated On: Jun 4, 2025
- \((e, e)\)
- \(\left(\dfrac{1}{e}, -1\right)\)
- \(\left(\dfrac{1}{e}, -\dfrac{2}{e^2}\right)\)
- \((e^3, 3e^3)\)
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The Correct Option is C
Solution and Explanation
Slope of line \(2x - 2y = 3 \Rightarrow y = x - \dfrac{3}{2} \Rightarrow m = 1\)
Normal to curve has slope \(m_n = 1\)
So tangent slope = \(-1\), as normal is perpendicular.
Find derivative: \[ y = x \log x \Rightarrow \dfrac{dy}{dx} = \log x + 1 = -1 \Rightarrow \log x = -2 \Rightarrow x = \dfrac{1}{e^2} \Rightarrow y = \dfrac{1}{e^2} \log \left( \dfrac{1}{e^2} \right) = \dfrac{1}{e^2} (-2) = -\dfrac{2}{e^2} \Rightarrow P = \left( \dfrac{1}{e}, -\dfrac{2}{e^2} \right) \]
Find derivative: \[ y = x \log x \Rightarrow \dfrac{dy}{dx} = \log x + 1 = -1 \Rightarrow \log x = -2 \Rightarrow x = \dfrac{1}{e^2} \Rightarrow y = \dfrac{1}{e^2} \log \left( \dfrac{1}{e^2} \right) = \dfrac{1}{e^2} (-2) = -\dfrac{2}{e^2} \Rightarrow P = \left( \dfrac{1}{e}, -\dfrac{2}{e^2} \right) \]
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