Q108 / 160 · TS EAMCET 2024 (10 May Shift 1) · Mathematics

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
$3 x-20 y+308=0$
Correct answer
D
$3 x+20 y=332$

Solution

$y^3=4 x^5 \Rightarrow \frac{d y}{d x}=\frac{20 x^4}{3 y^2} \Rightarrow\left(\frac{d y}{d x}\right)_{(4,16)}=\frac{20}{3}$ Slope of normal $=\frac{-3}{20}$ Equation of normal at $(4,16)$ and slope $\frac{-3}{20}$ is $y-16=\frac{-3}{20}(x-4) \Rightarrow 3 x+20 y=332$

Hence, the correct answer is option: 'D'.

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