The values of parameter a such that the line | vedclass

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(B) The given curve is $y = \frac{1}{x}$.$\frac{dy}{dx} = -\frac{1}{x^{2}}$.The slope of the tangent at any point $(x, y)$ is $-\frac{1}{x^{2}}$.The s

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$(0, 5)$

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(B) The given curve is $y = \frac{1}{x}$.$\frac{dy}{dx} = -\frac{1}{x^{2}}$.The slope of the tangent at any point $(x, y)$ is $-\frac{1}{x^{2}}$.The slope of the normal is the negative reciprocal of the tangent slope,which is $x^{2}$.Since $x^{2} > 0$ for all $x 
eq 0$,the slope of the normal must be positive.The given line is $(\log_{2}(1 + 5a - a^{2}))x - 5y - (a^{2} - 5) = 0$,which can be written as $y = \frac{\log_{2}(1 + 5a - a^{2})}{5}x - \frac{a^{2} - 5}{5}$.Thus,the slope of the line is $m = \frac{\log_{2}(1 + 5a - a^{2})}{5}$.Since $m > 0$,we have $\log_{2}(1 + 5a - a^{2}) > 0$.This implies $1 + 5a - a^{2} > 2^{0} = 1$.$5a - a^{2} > 0 \Rightarrow a^{2} - 5a $a(a - 5) < 0$,which gives $a \in (0, 5)$.

The values of parameter $a$ such that the line $(\log_{2}(1 + 5a - a^{2}))x - 5y - (a^{2} - 5) = 0$ is a normal to the curve $xy = 1$ may lie in the interval

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