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junior college 1 | H2 Maths
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Good morning!! How do i solve this question?
Date Posted: 4 years ago
Views: 412
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Remember that from your parametric eqn you can get (dy/dx) which is =(dy/dθ) ÷ (dx/dθ). So you know confirm somewhere you need to force out a gradient (also known as tangent) which can be found since the question gives you the gradient of the normal (normal // to y=x). So you derive the gradient of tangent and equate it to dy/dx and solve. Thereafter remember that question asking you for coordinates when the above happens which means need to sub in the θ into the y and x.
Date Posted:
4 years ago
Hmm, i did do it by your method but i got stuck after obtaining my Θ value (sin-1(-3/5)) bc idk how to proceed after subst. that value back into x and y
Okay I understand. Uploading another photo
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Sorry for the wait. For these parametric questions and in some differentiation/integration by u-substitution questions, always rely back on the drawing of trig triangle. There’s no way you can get “exact coordinates” if you use sin-1(-3/5). Also recognise the actual value of θ and put that into the given domain. In this case, θ is -36° which makes both sinθ and tanθ negative while cosθ is always positive for the given domain. So from sinθ=-3/5 you can derive cosθ and tanθ which you can then substitute into Y&X.
Date Posted:
4 years ago
I sub in the wrong value for the coordinate should be 5/cos θ not 5/sin θ but otherwise drawing of the trig triangle is still the main takeaway.
Ohh i see!! Thank you so so much! :D
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