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Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeTangents & NormalsMedium2 minQB

MathematicsMediumsingle choice

The equation of the normal to the curve parametrically represented by $x = t^{2} + 3 t - 8$ and $y = 2 t^{2} - 2 t - 5$ at the point $P (2, - 1)$ is

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$t^{2} + 3 t - 8 = 2 \Rightarrow t = 2, - 52 t^{2} - 2 t - 5 = - 1 \Rightarrow t = 2, - 1} \Rightarrow t = 2, \frac{d y}{d x} = \frac{4 t - 2}{2 t + 3} \Rightarrow (\frac{d y}{d x})_{t = 2} = \frac{6}{7}$ equation of normal: $y + 1 = \frac{- 7}{6} (x - 2) \Rightarrow 7 x + 6 y - 8 = 0$

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

⏱ 2mℹ️ Source: QB

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