Prove that the line is tangent to the curve at the point. Hello can someone please walk me through part a and b of the below question? I really want to understand it but am having a hard time figuring out the solution.
I know how to calculate curvature for a parametrized and unparametrized curve using the respective formulas. Would I just take the derivatives of alpha, the respective cross products for the formula and simplify down to find the curvature? 
Thank you very much. 
 A: It should be understood that the polar coordinates are for the tangent point T or $ \alpha$ and not for the foot of perpendicular P. 
I give a direct geometrical construction method based on differential lengths rather than a derivation.It is an attempt that is only partially successful,as of now, others may suggest improvements.  
Separate sketch labels $(r, t) $ polar coordinate tangency point T and also $\phi$ and $\psi$. 
From triangle OTX when external angle equated to sum of internal angles we get
 $$ t = \phi - \psi   ...(1*) $$
By consideration of differential triangles, we can mark differentials $ dL, dp, p d\theta$ and $ q dL. $ 
Also  $ d\theta = d \phi$ due to normal/tangent pair perpendicularity.
The radius r has components
$$ OP = p = r. sin \psi = dL/d\theta  ... (2*) $$ and,
$$ PT = q = -r. cos\psi = dp / d\theta  ...(3*) $$
along the normal and tangent respectively.
$$ x_T = p. cos\theta - q. sin \theta , y_T = p. sin\theta + q. cos \theta  ... (4*) $$ Proposition 7 a).
Derivative of (1*) with respect to arc results in curvature
$$ kg = d\psi/dL + sin \psi/r  ... (5*) $$
Plugging in from (2*) and (3*) 
$$ kg= dp/dL/q        ... (6*) $$
which does not agree with required result.
Arc Length $ L = \int p. d\theta $ Proposition 7 c).
but Area = $\int (p^2 + q^2)/2. d\theta $ which also does not agree with desired result.
EDIT2:
The given result in proposition 7 d), if it does connote area in another way, most probably is a printing error of the sign.

EDIT1:

The example of a circle radius a above x-axis may be considered as counter-example for Proposition 7 b).
$p = OV + VC + CT  = H/ sin \theta + h cos\theta  + a =  r\, sin \psi $
$ p' (\theta)  = - H\, csc \theta\,  cot \theta - h\, sin\theta $
$ p'' (\theta)  = - H\, csc\theta\,  ( csc^2\theta + cot^2\theta)  - h\, cos\theta  $
$  p  (\theta) +  p'' (\theta) = 2 H\, csc ^3\theta  + a $ 
which does not agree with 7 b) unless H=0.
