1.

If the curve y - ax3 = 4 and x2 = y, cut orthogonally at (-1, 1) then the value of a is 1. 12. \(\dfrac 12\)3. \(\dfrac 16\)4. None of these

Answer» Correct Answer - Option 3 : \(\dfrac 16\)

Concept:

If the tangent of the curves touches orthogonally then the product of their slopes = -1.

Calculations:

Given curves are y - ax3 = 4 and x2 = y

slope of tangent of the curve y - ax3  = 4.is \(\rm m_1 = \dfrac {dy}{dx} = 3x^2a\)

⇒ slope of tangent of the curve y - ax3  = 4 at the point (- 1, 1) is \(\rm m_1 =\) 3a.

Now, slope of tangent of the curve x2 = y is \(\rm m_2 = \dfrac {dy}{dx} = 2x\)

⇒ slope of tangent of the curve x2 = y at the point (- 1, 1) is \(\rm m_2 = - 2\) 

Given, the curve y - ax3 = 4 and x2 = y, cut orthogonally at (-1, 1).

⇒ \(\rm m_1 .m_ 2 = -1\)

⇒ (3a)(-2) = - 1

⇒ a = \(\dfrac 16\)


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