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Let $x =2 \sqrt{2}\cos\theta, \; y =2 \sqrt{2}\sin\theta$ and $x =2 t, \; y =\frac{2}{t}$ are two curves then there curves in first quadrant

Option 1)
intersect at $\frac{\pi}{6}$

Option 2)
Touch each other

Option 3)
intersect at $\frac{\pi}{}4$

Option 4)
intersect at $\frac{\pi}{}3$

Answers (1)

best_answer

As we have learened

Condition for the two curves to touch in parametric form -

$\frac{g'_{x}}{g'_{y}}=\frac{f'_{x}}{f'_{y}}$

- wherein

Where

$x=f(t)$

$y=f(t)$

For point of intersection or touching i.e common point : $2\sqrt2 \cos \theta =2t$ and

$2\sqrt2 \sin \theta =2/t$

$\Rightarrow t= \sqrt2 \cos \theta$ and $2\sqrt2 \sin \theta \times \sqrt2 \cos \theta = 2\Rightarrow \sin 2\theta$

$\Rightarrow \theta = \pi /4$ for first quadrant common point and t =1

$\therefore$ point is (2,2)

$\frac{fx^{1}}{fy^{1}}= \frac{-2 \sqrt 2 \sin \theta }{ 2\sqrt 2 \cos\theta }= - \tan \theta \Rightarrow \frac{fx^{1}}{fy^{1}}$ at $\theta = \pi /4$ is -1

$\frac{gx^{1}}{gy^{1}}= \frac{2}{-2/t^{2}}= -t^{2}\Rightarrow \frac{gx^{1}}{gy^{1}}$ at t= 1 is -1

$\frac{fx^{1}}{fy^{1}}=\frac{gx^{1}}{gy^{1}}$ so curves touch each other at common point

Option 1)

intersect at $\frac{\pi}{6}$

Option 2)

Touch each other

Option 3)

intersect at $\frac{\pi}{}4$

Option 4)

intersect at $\frac{\pi}{}3$

Posted by

Himanshu

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