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The distance, from the origin, of the normal to the curve,

$x= 2\cos \! t+2t\sin \! t, y=2\sin \! t-2t\cos \! t\: \: at\; \; t= \frac{\pi }{4}$ is

Option 1)
$4$

Option 2)
$2\sqrt{2}$

Option 3)
$2$

Option 4)
$\sqrt{2}$

Answers (1)

best_answer

As we learnt in

Perpendicular distance of a point from a line -

$\rho =\frac{\left | ax_{1}+by_{1}+c\right |}{\sqrt{a^{2}+b^{2}}}$

- wherein

$\rho$ is the distance from the line $ax+by+c=0$ .

Slope – point from of a straight line -

$y-y_{1}=m(x-x_{1})$

- wherein

$m\rightarrow$ slope

$\left ( x_{1},y_{1} \right )\rightarrow$ point through which line passes

$x=2\cos t+2t\sin t$

$y=2\sin t-2t\cos t$

at $t=\frac{\pi }{4}$

$x=\sqrt{2}+\frac{\sqrt {2}\pi }{4}$ ; $y=\sqrt{2}-\frac{\sqrt{2}\pi }{4}$

$\left. \frac{dy}{dx} \right \}_{t=\frac{\pi}{4}}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ $=\frac{2t\cos t}{+2t\sin t}=\cot \left ( t \right )$

Slope of normal = $=\left. -tan^{-1}(t) \right \]_{\frac{\pi}{4}}=-1$

equation of normal

$\Rightarrow y-\left ( \sqrt{2}-{\frac{\sqrt{2}\pi }{4}} \right )=-1\left (x- \sqrt{2}-{\frac{\sqrt{2}\pi }{4}} \right )$

$x+y=2\sqrt{2}$

Distance from origin= $\frac{\left ( 0+0-2\sqrt{2} \right )}{\sqrt{1^{2}+1^{2}}}$

= 2 units

Option 1)

$4$

Incorrect option

Option 2)

$2\sqrt{2}$

Incorrect option

Option 3)

$2$

Correct option

Option 4)

$\sqrt{2}$

Incorrect option

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prateek

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