If the normal drawn to the hyperbola xy=16 at | vedclass

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(B) The equation of the hyperbola is $xy = 16$. The slope of the tangent at $(x_1, y_1)$ is given by $\frac{dy}{dx} = -\frac{y}{x}$. At $(8, 2)$,the s

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$34$

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(B) The equation of the hyperbola is $xy = 16$. The slope of the tangent at $(x_1, y_1)$ is given by $\frac{dy}{dx} = -\frac{y}{x}$. At $(8, 2)$,the slope $m = -\frac{2}{8} = -\frac{1}{4}$.The slope of the normal is $m_n = -\frac{1}{m} = 4$.The equation of the normal at $(8, 2)$ is $y - 2 = 4(x - 8)$,which simplifies to $y = 4x - 30$.To find the intersection with the hyperbola,substitute $y = 4x - 30$ into $xy = 16$:$x(4x - 30) = 16 \implies 4x^2 - 30x - 16 = 0 \implies 2x^2 - 15x - 8 = 0$.Factoring the quadratic: $(2x + 1)(x - 8) = 0$.The roots are $x = 8$ (the original point) and $x = -\frac{1}{2}$.For $x = \alpha = -\frac{1}{2}$,we find $\beta = \frac{16}{\alpha} = \frac{16}{-1/2} = -32$.We need to calculate $|\beta| + \frac{1}{|\alpha|} = |-32| + \frac{1}{|-1/2|} = 32 + 2 = 34$.

If the normal drawn to the hyperbola $xy=16$ at $(8,2)$ meets the hyperbola again at a point $(\alpha, \beta)$,then $|\beta|+\frac{1}{|\alpha|}=$

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