MHT CET · Maths · Application of Derivatives
The equation of the normal to the curve \(2 x^{2}+y^{2}=12\) at the point \((2,2)\) is
- A \(2 x-y+6=0\)
- B \(2 x+y-6=0\)
- C \(x+2 y+2=0\)
- D \(x-2 y+2=0\)
Answer & Solution
Correct Answer
(D) \(x-2 y+2=0\)
Step-by-step Solution
Detailed explanation
(A)
Given equation of the curve is \(2 x^{2}+y^{2}=12\)
\(\therefore \quad 4 x+2 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{-2 x}{y}\)
Slope of tangent at \((2,2)\) is \(\frac{-2(2)}{2}-2 \Rightarrow\) Slope of normal at \((2,2)\) is \(\frac{1}{2}\)
Equation of normal at \((2,2)\) is
\(y-2=\frac{1}{2}(x-2) \Rightarrow 2 y-4=x-2 \Rightarrow\) \(x-2 y+2=0\)
Given equation of the curve is \(2 x^{2}+y^{2}=12\)
\(\therefore \quad 4 x+2 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{-2 x}{y}\)
Slope of tangent at \((2,2)\) is \(\frac{-2(2)}{2}-2 \Rightarrow\) Slope of normal at \((2,2)\) is \(\frac{1}{2}\)
Equation of normal at \((2,2)\) is
\(y-2=\frac{1}{2}(x-2) \Rightarrow 2 y-4=x-2 \Rightarrow\) \(x-2 y+2=0\)
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