Find the family of curves such that the angle | vedclass

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(D) Let $m = \frac{dy}{dx}$ be the slope of the tangent to the required curve at $(x, y)$.The slope of the tangent to the curve $xy = c^2$ at $(x, y)$

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All of the above

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(D) Let $m = \frac{dy}{dx}$ be the slope of the tangent to the required curve at $(x, y)$.The slope of the tangent to the curve $xy = c^2$ at $(x, y)$ is $m_1 = -\frac{y}{x}$.The angle between the two tangents is $\frac{\pi}{4}$,so $\left| \frac{m - m_1}{1 + m m_1} ight| = 	an\left( \frac{\pi}{4} ight) = 1$.Substituting $m_1 = -\frac{y}{x}$,we get $\left| \frac{m + \frac{y}{x}}{1 - m \frac{y}{x}} ight| = 1$,which implies $\frac{m + \frac{y}{x}}{1 - m \frac{y}{x}} = \pm 1$.Case $1$: $m + \frac{y}{x} = 1 - m \frac{y}{x} \implies m(1 + \frac{y}{x}) = 1 - \frac{y}{x} \implies m = \frac{x-y}{x+y}$. This is a homogeneous differential equation leading to $y^2 + 2xy - x^2 = k$.Case $2$: $m + \frac{y}{x} = -(1 - m \frac{y}{x}) \implies m(1 - \frac{y}{x}) = -1 - \frac{y}{x} \implies m = \frac{x+y}{x-y}$. This leads to $y^2 - 2xy - x^2 = k$.Since both solutions are valid,the correct option is $D$.

Find the family of curves such that the angle between the tangent at any point $(x, y)$ and the tangent to the curve $xy = c^2$ at the point of intersection is $\frac{\pi}{4}$.

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