The angle between the tangents drawn at (0,0) | vedclass

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Question

(D) For the curve $y^3-x^2 y+5 y-2 x=0$,differentiating with respect to $x$ gives: $3 y^2 \frac{dy}{dx} - (x^2 \frac{dy}{dx} + 2xy) + 5 \frac{dy}{dx}

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$\frac{\pi}{2}$

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(D) For the curve $y^3-x^2 y+5 y-2 x=0$,differentiating with respect to $x$ gives: $3 y^2 \frac{dy}{dx} - (x^2 \frac{dy}{dx} + 2xy) + 5 \frac{dy}{dx} - 2 = 0$ At $(0,0)$,this simplifies to $5 \frac{dy}{dx} - 2 = 0$,so the slope $m_1 = \frac{2}{5}$. For the curve $x^4-x^3 y^2+5 x+2 y=0$,differentiating with respect to $x$ gives: $4x^3 - (x^3 \cdot 2y \frac{dy}{dx} + 3x^2 y^2) + 5 + 2 \frac{dy}{dx} = 0$ At $(0,0)$,this simplifies to $5 + 2 \frac{dy}{dx} = 0$,so the slope $m_2 = -\frac{5}{2}$. Since $m_1 	imes m_2 = (\frac{2}{5}) 	imes (-\frac{5}{2}) = -1$,the tangents are perpendicular. Therefore,the angle between them is $\frac{\pi}{2}$.

The angle between the tangents drawn at $(0,0)$ to the curves $y^3-x^2 y+5 y-2 x=0$ and $x^4-x^3 y^2+5 x+2 y=0$ is

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