MHT CET · Maths · Application of Derivatives
The point of the curve \(y^{2}=2(x-3)\) at which the normal is parallel to the line \(y-2 x+1=0\) is
- A \((5,2)\)
- B \(\left(-\frac{1}{2},-2\right)\)
- C \((5,-2)\)
- D \(\left(\frac{3}{2}, 2\right)\)
Answer & Solution
Correct Answer
(C) \((5,-2)\)
Step-by-step Solution
Detailed explanation
Given, \(y^{2}=2(x-3) \quad\)...(i)
On differentiating w.r.t. \(x\), we get \(2 y \frac{d y}{d x}=2\)
\(\Rightarrow\)
\(\frac{d y}{d x}=\frac{1}{y}\)
Slope of the normal \(=\frac{-1}{(d y / d x)}=-y\) Slope of the given line \(=2\) \(\therefore \quad y=-2\)
From Eq. (i), \(x=5\)
\(\therefore\) Required point is \((5,-2)\).
On differentiating w.r.t. \(x\), we get \(2 y \frac{d y}{d x}=2\)
\(\Rightarrow\)
\(\frac{d y}{d x}=\frac{1}{y}\)
Slope of the normal \(=\frac{-1}{(d y / d x)}=-y\) Slope of the given line \(=2\) \(\therefore \quad y=-2\)
From Eq. (i), \(x=5\)
\(\therefore\) Required point is \((5,-2)\).
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