The curve $x^4-2 x y^2+y^2+3 x-3 y=0$ cuts the X -axis at $(0,0)$ at an angle of

The curve $x^4-2 x y^2+y^2+3 x-3 y=0$ cuts the X -axis at $(0,0)$ at an angle of
  1. $\frac{\pi}{4}$
  2. $\frac{\pi}{2}$
  3. 0
  4. $\frac{\pi}{6}$

Solution

Given equation of curve is $x^4-2 x y^2+y^2+3 x-3 y=0...(i)$ Slope of tangent $=m=\frac{d y}{d x}$ at $(0,0)$ Differentiating (i) w.r.to $x$, we get $\begin{array}{ll} & 4 x^3-2 x \cdot 2 y \frac{\mathrm{~d} y}{\mathrm{~d} x}-y^2(2)+2 y \frac{\mathrm{~d} y}{\mathrm{~d} x}+3-3 \frac{\mathrm{~d} y}{\mathrm{~d} x}=0 \\ \therefore & \quad \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{4 x^3-2 y^2+3}{4 x y-2 y+3} \\ \therefore \quad & \mathrm{~m}=\left.\frac{\mathrm{d} y}{\mathrm{~d} x}\right|_{(0,0)}=\frac{4(0)-2(0)+3}{4(0) \cdot(0)-2(0)+3}=\frac{3}{3}=1 \\ \therefore \quad & \mathrm{~m}=1 \\ \therefore \quad & \tan \theta=1 \\ \therefore \quad & \theta=\frac{\pi}{4} \end{array}$

Asked in: MHT CET 2024 (10 May Shift 1)

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