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)

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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