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STD 11 - 10.2 parabola,ellipse,hyperbola — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 10.2 parabola,ellipse,hyperbola4 Marks
MCQ
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
    $\left( { - \infty ,0} \right)$
  • ✓
    $(0, 5)$
  • C
    $(5, 10)$
  • D
    $\left( {10,\infty } \right)$
✓

Answer

Correct option: B.
$(0, 5)$
b
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 $a$ $\in(0,5)$

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