Q. 3 B5.0( 1 Vote )

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Answer :

Given:


Curves x2 = y ...(1)


& x3 + 6y = 7 ...(2)


The point of intersection of two curves (1,1)


Solving (1) & (2),we get,


First curve is x2 = y


Differentiating above w.r.t x,


2x



m1 = 2x ...(3)


Second curve is x3 + 6y = 7


Differentiating above w.r.t x,


3x2 + 6. = 0




m2 ...(4)


Substituting (1,1) for m1 & m2,we get,


m1 = 2x


2×1


m1 = 2 ...(5)


m2



m2 = ...(6)


when m1 = 2 & m2 =



1


Two curves x2 = y & x3 + 6y = 7 intersect orthogonally.


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