Parametric Help

Attached. Done part A by differentiating via chain rule, and getting values by substituting t. got 2/3 and -2/3.

part b used my answer to part a to form equations in form y-y1=m(x-x1)

Answer isn't the same line though...

Could someone also walk me through how you would work out part c?

you know it meets y axis therefore x =0. sub x=0 into equation of tangent. However could I work this out by just subtituting values of t into original equations then I will know the coordinates and can work out the area from there?

I get $y=-\frac{2}{3}x+1$ for both lines.



Bit puzzled by this. Question says tangent meets the y-axis at Q, but Q is one of our points and it's not on the y-axis. So, it's not clear which triangle they're talking about.

Can u show it at P for part b? since that's the one that doesn't have the answer. I'm having a problem with the root 2, cause it's the 'wrong sign' however mathematically it isn't in my method.

At P, t=pi/4, and eqn of tgt is:

$y-(1-\frac{2}{\sqrt{2}})=-\frac{2}{3}(x-\frac{3}{\sqrt{2}})$

oh sorry, meant Q then, whichever one where t=5pi/4. I got that one and got it to equal y=-2/3 x +1