मराठी

If the Curve Ay + X2 = 7 and X3 = Y Cut Orthogonally at (1, 1), Then a is Equal to (A) 1 (B) −6 (C) 6 (D) 0 - Mathematics

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प्रश्न

If the curve ay + x2 = 7 and x3 = y cut orthogonally at (1, 1), then a is equal to _____________ .

पर्याय

  • 1

  • `-6`

  • 6

  • 0

MCQ
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उत्तर

`6`

 

\[\text { Given }: \]

\[ay + x^2 = 7 . . . \left( 1 \right)\]

\[ x^3 = y . . . \left( 2 \right)\]

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

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

\[a\frac{dy}{dx} + 2x = 0\]

\[ \Rightarrow \frac{dy}{dx} = \frac{- 2x}{a}\]

\[ \Rightarrow m_1 = \left( \frac{dy}{dx} \right)_\left( 1, 1 \right) = \frac{- 2}{a}\]

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

\[3 x^2 = \frac{dy}{dx}\]

\[ \Rightarrow m_2 = \left( \frac{dy}{dx} \right)_\left( 1, 1 \right) = 3\]

\[\text { It is giventhat the curves are orthogonal at the given point }.\]

\[ \therefore m_1 \times m_2 = - 1\]

\[ \Rightarrow \frac{- 2}{a} \times 3 = - 1\]

\[ \Rightarrow a = 6\]

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पाठ 16: Tangents and Normals - Exercise 16.5 [पृष्ठ ४३]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 16 Tangents and Normals
Exercise 16.5 | Q 15 | पृष्ठ ४३

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