MHT CET · Maths · Application of Derivatives
The point on the curve \(y^2=2(x-3)\) at which the normal is parallel to the line \(y-2 x+1=0\) is
- A \(\left(\frac{-1}{2},-2\right)\)
- B \(\left(\frac{3}{2}, 2\right)\)
- C \((5,2)\)
- D \((5,-2)\)
Answer & Solution
Correct Answer
(D) \((5,-2)\)
Step-by-step Solution
Detailed explanation
\(y^2=2(x-3) \)
\( \therefore 2 y \frac{d y}{d x}=2 \quad \Rightarrow \frac{d y}{d x}=\frac{1}{y} \)
\( \therefore \text { Slope of normal}=-y \text { and as per condition given} \)
\(-y=2 \quad \Rightarrow y=-2 \)
\( \therefore \quad(-2)^2=2(x-3) \Rightarrow x=5 \Rightarrow \text { point is }(5,-2)\)
\( \therefore 2 y \frac{d y}{d x}=2 \quad \Rightarrow \frac{d y}{d x}=\frac{1}{y} \)
\( \therefore \text { Slope of normal}=-y \text { and as per condition given} \)
\(-y=2 \quad \Rightarrow y=-2 \)
\( \therefore \quad(-2)^2=2(x-3) \Rightarrow x=5 \Rightarrow \text { point is }(5,-2)\)
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