Abscissae of points on the curve xy = (c + x) | vedclass

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(C) Given the curve $xy = (x + c)^2$,we can write $y = \frac{(x + c)^2}{x} = \frac{x^2 + 2cx + c^2}{x} = x + 2c + \frac{c^2}{x}$.Differentiating with

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$\pm c / \sqrt{2}$

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(C) Given the curve $xy = (x + c)^2$,we can write $y = \frac{(x + c)^2}{x} = \frac{x^2 + 2cx + c^2}{x} = x + 2c + \frac{c^2}{x}$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = 1 - \frac{c^2}{x^2} = \frac{x^2 - c^2}{x^2}$.The slope of the normal at any point $(x, y)$ is $m_n = -\frac{dx}{dy} = -\frac{x^2}{x^2 - c^2} = \frac{x^2}{c^2 - x^2}$.The normal cuts off numerically equal intercepts from the axes,which means the slope of the normal must be $\pm 1$.Case $1$: $\frac{x^2}{c^2 - x^2} = 1 \Rightarrow x^2 = c^2 - x^2 \Rightarrow 2x^2 = c^2 \Rightarrow x = \pm \frac{c}{\sqrt{2}}$.Case $2$: $\frac{x^2}{c^2 - x^2} = -1 \Rightarrow x^2 = -(c^2 - x^2) \Rightarrow x^2 = -c^2 + x^2 \Rightarrow 0 = -c^2$,which is impossible for $c 
eq 0$.Thus,the abscissae are $\pm \frac{c}{\sqrt{2}}$.

Abscissae of points on the curve $xy = (c + x)^2$,the normal at which cuts off numerically equal intercepts from the axes of coordinates is/are:

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