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At x =t,\; y =\frac{\sqrt{2}}{t} and x =2t_1^2,\; y =2t_1 are two curves f(x) & g(x). Which inetrsect at some angle \theta, then their intersection 

  • Option 1)

    is orthogonal

  • Option 2)

    make \theta = \frac{\pi}{}4

  • Option 3)

    make \theta = \frac{\pi}{}6

  • Option 4)

    make \theta = \frac{\pi}{}3

 

Answers (1)

best_answer

As we have learned 

Condition of Orthogonality in parametric form -

f'_{x}.g'_{x}+f'_{y}.g'_{y}=0

- wherein

Where

x=f(t)

y=f(t)

 

 For point of intersection :- t= 2t_{1}^2{}  and \sqrt{2}/ t = 2t_{1}

\Rightarrow \frac{\sqrt2}{2t_{1}^{2}}=2t_{1} = 1/\sqrt 2\Rightarrow 1

\therefore  point is (1, \sqrt{2} )

f'(x)\cdot g'(x)+ f'(y)\cdot g'(y)= 1*4t_{1}+ \left ( \frac{-\sqrt2 }{t^{2}} \right )(2)

= 4(1/\sqrt 2) -2\sqrt 2 (1/1) =0

\therefore Both curves intersect orthogonally 

 

 

 

 

 


Option 1)

is orthogonal

Option 2)

make \theta = \frac{\pi}{}4

Option 3)

make \theta = \frac{\pi}{}6

Option 4)

make \theta = \frac{\pi}{}3

Posted by

Himanshu

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