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

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(-3,3)

(0,3)

(1,3)

$(- \infty, 0) \cup (3, \infty)$

answer is A.

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Detailed Solution

$xy = 1 \Rightarrow y = \frac{1}{x}$
Given $\frac{dy}{dx} = \frac{- 1}{x^{2}}$
$slope of \tan gent is < 0 \Rightarrow slope of normal > 0$
$\Rightarrow \frac{- (3 - a)}{a} > 0 \Rightarrow a \in (- \infty, 0) \cup (3, \infty)$

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