Question

The value of the parameter ${}^{\prime }{a}^{\prime }$ so that the line $(3-a)x+ay+(a^2-1)=0$ is a normal to the curve $xy=1,$ may lies in the interval

• A. $(-\infty ,0)\cup (3,\infty )$
• B. $(1,3)$
• C. $(-3,3)$
• D. $(0,3)$

Answer 1

The correct option is A $(-\infty ,0)\cup (3,\infty )$

$\frac{dy}{dx}=-\frac{1}{x^2}$
$\Rightarrow$ Slope of tangent $<0$
$\therefore$ Slope of normal $>0$
$\Rightarrow \frac{a-3}{a}>0$
So, $a\in (-\infty ,0)\cup (3,\infty )$