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The Angle Between the Curves Y2 = X and X2 = Y at (1, 1) is - Mathematics

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Question

The angle between the curves y2 = x and x2 = y at (1, 1) is ______________ .

Options

  • \[\tan^{- 1} \frac{4}{3}\]

  • \[\tan^{- 1} \frac{3}{4}\]

  • 90°

  • 45°

MCQ
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Solution

\[\tan^{- 1} \frac{3}{4}\]

 

\[\text { Given }: \]

\[ y^2 = x . . . \left( 1 \right)\]

\[ x^2 = y . . . \left( 2 \right)\]

\[\text { Point} = \left( 1, 1 \right)\]

\[\text { On differentiating (1) w.r.t. x, we get }\]

\[2y \frac{dy}{dx} = 1\]

\[ \Rightarrow \frac{dy}{dx} = \frac{1}{2y}\]

\[ \Rightarrow m_1 = \frac{1}{2}\]

\[\text { On differentiating (2) w.r.t.x, we get }\]

\[2x = \frac{dy}{dx}\]

\[ \Rightarrow m_2 = 2\left( 1 \right) = 2\]

\[\text { Now,} \]

\[\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{\frac{1}{2} - 2}{1 + \frac{1}{2} \times 2} \right| = \frac{3}{4}\]

\[ \Rightarrow \theta = \tan^{- 1} \left( \frac{3}{4} \right)\]

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Chapter 16: Tangents and Normals - Exercise 16.5 [Page 42]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 16 Tangents and Normals
Exercise 16.5 | Q 9 | Page 42

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