The tangent of the angle between the curves x | vedclass

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(C) Given curves are $xy=1$ $(i)$ and $x^2+8y=0$ (ii). From $(i)$, $y = \frac{1}{x}$. Substituting in (ii): $x^2 + 8(\frac{1}{x}) = 0 \Rightarrow x^3

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$\frac{6}{7}$

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(C) Given curves are $xy=1$ $(i)$ and $x^2+8y=0$ (ii). From $(i)$, $y = \frac{1}{x}$. Substituting in (ii): $x^2 + 8(\frac{1}{x}) = 0 \Rightarrow x^3 + 8 = 0 \Rightarrow x^3 = -8 \Rightarrow x = -2$. For $x = -2$, $y = \frac{1}{-2} = -\frac{1}{2}$. The point of intersection is $(-2, -\frac{1}{2})$. Differentiating $(i)$: $y + x \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$. At $(-2, -\frac{1}{2})$, $m_1 = -\frac{-1/2}{-2} = -\frac{1}{4}$. Differentiating (ii): $2x + 8 \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{4}$. At $(-2, -\frac{1}{2})$, $m_2 = -\frac{-2}{4} = \frac{1}{2}$. The tangent of the angle $	heta$ between the curves is given by $	an 	heta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$. Substituting the values: $	an 	heta = |\frac{-1/4 - 1/2}{1 + (-1/4)(1/2)}| = |\frac{-3/4}{1 - 1/8}| = |\frac{-3/4}{7/8}| = |-\frac{3}{4} 	imes \frac{8}{7}| = |-\frac{6}{7}| = \frac{6}{7}$.

The tangent of the angle between the curves $xy=1$ and $x^2+8y=0$ is

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