If the point on y = x tan α − 2 u 2 c o s 2 α a x 2 ( 0 < α < 2 π ) where the tangent is parallel to y = x has an ordinate 4 a u 2 then the value of α is

Given m = 1 ⇒ tan α − u 2 c o s 2 α a x = 1 ⇒ x = a ( t a n α − 1 ) u 2 cos 2 α Substitute x in the given equation: y = x ( tan α − 2 t a n α − 1 ) = 2 a u 2 ( sin 2 α − cos 2 α ) = − 2 a u 2 cos 2 α 4 a u 2 = − 2 a u 2 cos 2 α ⇒ cos 2 α = − 2 1 ⇒ α = 3 π

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MathematicsApplication of DerivativeTangents & NormalsMedium2 minQB

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If the point on $y = x tan α - \frac{a x ^{2}}{2 u ^{2} c o s ^{2} α} (0 < α < \frac{π}{2})$ where the tangent is parallel to $y = x$ has an ordinate $\frac{u ^{2}}{4 a}$ then the value of $α$ is

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Given $m = 1 \Rightarrow tan α - \frac{a x}{u ^{2} c o s ^{2} α} = 1 \Rightarrow x = \frac{( t a n α - 1 )}{a} u^{2} cos^{2} α$ Substitute x in the given equation: $y = x (tan α - \frac{t a n α - 1}{2}) = \frac{u ^{2}}{2 a} (sin^{2} α - cos^{2} α) = - \frac{u ^{2}}{2 a} cos 2 α$ $\frac{u ^{2}}{4 a} = - \frac{u ^{2}}{2 a} cos 2 α \Rightarrow cos 2 α = - \frac{1}{2} \Rightarrow α = \frac{π}{3}$

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

⏱ 2mℹ️ Source: QB

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