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

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
\(4 x^3-2 x \cdot 2 y \frac{\mathrm{~d} y}{\mathrm{~d} x}-y^2(2)+2 y \frac{\mathrm{~d} y}{\mathrm{~d} x}+3-3\) \( \frac{\mathrm{~d} y}{\mathrm{~d} x}=0 \)
\( \therefore \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{4 x^3-2 y^2+3}{4 x y-2 y+3} \)
\( \therefore \mathrm{~m}=\left.\frac{\mathrm{d} y}{\mathrm{~d} x}\right|_{(0,0)}=\frac{4(0)-2(0)+3}{4(0) \cdot(0)-2(0)+3}=\frac{3}{3}=1 \)
\( \therefore \mathrm{~m}=1 \)
\( \therefore \tan \theta=1 \)
\( \therefore \theta=\frac{\pi}{4}\)