Values of parameter 'a' such that line left( {{{log }₂}left( {1 + 5a - {a^2}} right)} righ

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10-2. Parabola, Ellipse, Hyperbola

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The values of parameter $'a'$ such that the line $\left( {{{\log }_2}\left( {1 + 5a - {a^2}} \right)} \right)x - 5y - \left( {{a^2} - 5} \right) = 0$ is a normal to the curve $xy = 1$ , may lie in the interval

A

$\left( { - \infty ,0} \right)$

B

$(0, 5)$

C

$(5, 10)$

D

$\left( {10,\infty } \right)$

Solution

Given curve is $y=\frac{1}{x} $

$\Rightarrow \frac{d y}{d x}=-\frac{1}{x^{2}}$

$\therefore $ slope of normal $\left.=x^{2}>0 \text { (As } x \neq 0\right)$

$\therefore $ slope of given line $=\frac{\log _{2}\left(1+5 a-a^{2}\right)}{5}>0$

$\Rightarrow \log _{2}\left(1+5 \mathrm{a}-\mathrm{a}^{2}\right)>0 $

$\Rightarrow 1+5 \mathrm{a}-\mathrm{a}^{2}>1$

$\Rightarrow a^{2}-5 a<0$

Hence $a$ $\in(0,5)$

Standard 11

Mathematics

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