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Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeTangents & NormalsMedium2 minQB

MathematicsMediummultiple choice

The values of the parameter 'a' so that the line $(3 - a) x + a y + a^{2} - 1 = 0$ is a normal to the curve $x y = 1$ may belong to the interval

Options:(select one or more)

Answer:

A, B

Solution:

Slope of given line $= \frac{a - 3}{a}$ . For $x y = 1, \frac{d y}{d x} = - \frac{1}{x ^{2}}$ , so slope of normal $= x^{2} > 0$ . $∴ \frac{a - 3}{a} > 0 \Rightarrow a > 3$ or $a < 0$ , i.e. $a \in (3, \infty) \cup (- \infty, 0)$

Stream:JEESubject:MathematicsTopic:Application of DerivativeSubtopic:Tangents & Normals

⏱ 2mℹ️ Source: QB

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