TS EAMCET · Maths · Application of Derivatives
The equation of the normal drawn to the curve \(y^3=4 x^5\) at the point \((4,16)\) is
- A \(20 x+3 y=128\)
- B \(20 x-3 y=32\)
- C \(3 x-20 y+308=0\)
- D \(3 x+20 y=332\)
Answer & Solution
Correct Answer
(D) \(3 x+20 y=332\)
Step-by-step Solution
Detailed explanation
\(y^3=4 x^5 \Rightarrow \frac{d y}{d x}=\frac{20 x^4}{3 y^2} \Rightarrow\left(\frac{d y}{d x}\right)_{(4,16)}=\frac{20}{3}\) Slope of normal \(=\frac{-3}{20}\) Equation of normal at \((4,16)\) and slope \(\frac{-3}{20}\) is \(y-16=\frac{-3}{20}(x-4) \Rightarrow 3 x+20 y=332\)
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