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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 $\theta = \frac{\pi}{}3$

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Himanshu

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