WBJEE · Maths · Application of Derivatives
If the tangent at the point \(P\) with co-ordinates \((h, k)\) on the curve \(y^{2}=2 x^{3}\) is perpendicular to the straight line \(4 x=3 y\), then
- A \((\mathrm{h}, \mathrm{k})=(0,0)\) only
- B \((\mathrm{h}, \mathrm{k})=\left(\frac{1}{8},-\frac{1}{16}\right)\) only
- C \((\mathrm{h}, \mathrm{k})=(0,0)\) or \(\left(\frac{1}{8},-\frac{1}{16}\right)\)
- D no such point P exists
Answer & Solution
Correct Answer
(B) \((\mathrm{h}, \mathrm{k})=\left(\frac{1}{8},-\frac{1}{16}\right)\) only
Step-by-step Solution
Detailed explanation
\(2 y \frac{d y}{d x}=6 x^{2} \Rightarrow \frac{d y}{d x}=\frac{3 x^{2}}{y}=-\frac{3}{4} \Rightarrow y=-4 x^{2}\) and \(y^{2}=2 x^{3}\) \(\Rightarrow 16 x^{4}=2 x^{3} \Rightarrow x=\frac{1}{8} \quad(x \neq 0)\) \(y=\frac{-1}{16} \quad(y \neq 0)\)
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