I have tried this question more than a dozen times now. The question is find the point(s) at which the line normal to $y = 2\arcsin 0.5x$ is parallel to the line $y = 1-x$.

Edit: I changed y= 2arcsin0.5x to y=1/2arcsin0.5x then derive it. Which is wrong. Thanks for the answers.

  • $\begingroup$ have you tried doing it? $\endgroup$ – Pushkar Soni Jan 22 '17 at 4:47
  • $\begingroup$ Yes I did and i wasn't doing the right steps. Check edit. $\endgroup$ – n.j Jan 22 '17 at 5:11
  • $\begingroup$ just do the same procedure, you'll get the required equation. $\endgroup$ – Pushkar Soni Jan 22 '17 at 5:15
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    $\begingroup$ however, the site is not meant for the purpose of homework questions. when you ask a question please provide the solution you have done. if you are having a problem with a particular step. people will help, otherwise not. and please do not expect complete answers to these questions this site will only provide you a hint. so you could solve it yourself. $\endgroup$ – Pushkar Soni Jan 22 '17 at 5:20
  • $\begingroup$ i guess you are having problem regarding differentiation.right? $\endgroup$ – Pushkar Soni Jan 22 '17 at 5:22

we have, $$ y= arcsin(0.5)$$ differentiate it to get

$$ \frac{1}{\sqrt{(1-0.25x^2)}} $$ which is ofcourse the slope of the tangent.

then the slope of the normal is given by:

$$ -\frac{\sqrt{(1-0.25x^2)}}{1} $$

for second equation


the slope is $$-1$$

according to your question both the slope of normal and slope of the line should be equal since they are parallel.

equate both the equation to get the required result.



$$y=2\arcsin 0.5 x$$

$$\frac{dy}{dx}=\frac{2(0.5)}{\sqrt{1-(0.5x)^2}} $$

The gradient of the normal line would be $-\sqrt{1-(0.5x)^2}$


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