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Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeTangents & NormalsMedium2 minQB

MathematicsMediumsingle choice

The equation of the straight lines which are both tangent and normal to the curve $27 x^{2} = 4 y^{3}$ are

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Solution:

$x = 2 t^{3}, y = 3 t^{2} \Rightarrow$ tangent at t is $x - y t = - t^{3}$ . Normal at $t_{1}$ is, $x t_{1} + y = 2 t_{1}^{4} + 3 t_{1}^{2}$ $\Rightarrow \frac{1}{t _{1}} = - t = \frac{- t ^{3}}{2 t _{1}^{4} + 3 t _{1}^{2}} \Rightarrow t^{6} - 3 t^{2} - 2 = 0 \Rightarrow t^{2} = 2 \Rightarrow t = \pm 2$ ∴ lines are $x = \pm 2 (y - 2)$

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

⏱ 2mℹ️ Source: QB

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