$r$
is tangent at two points on the parabola $y=x^2$
such that the angle between the two radii at the tangent points is $2\theta$
, where $0<2\theta<\pi$
. Find $r$
as a function of $\theta$
.

 What is the slope of the tangent to the curve
$y = 2x^2+3x2$
at$x=4$
?  Find the value of
$a$
such that the line$y_1=3x+a$
is tangent to the curve$y_2=2x^2+3x+1$
 You are given
$3$
lines:$y=2x$
,$y=2x2$
,$y=\frac{1}{2}x+3$
. For each pair of lines, state whether they are parallel or perpendicular.
 What is the slope of the tangent to the curve

Let
$P$
be one of the points where the parabola and the circle touch. What is the slope of the tangent at$P$
, given you know the equation of the curve? 
What is the angle made by the tangent at
$P$
with the$x$
axis in terms of$\theta$
? 
What is another expression of the slope of the tangent at
$P$
in terms of this angle. 
Can you extract the tangent of this angle from the right triangle with sides
$r,$
$x$
and$r\cos(\theta)?$

Let
$P=(x,x^2)$
be one of the two points where the parabola and circle are tangent to each other. The radius$r$
sits on the line normal at$P$
. The slope of the tangent line at$P$
is equal to the derivative of$x^2$
, i.e.$2x$
, but also equal to the tangent of the angle the line makes with the$x$
axis, which in this case is$\theta$
(owing to the fact we have a right angle at$P$
between the tangent and the normal), i.e.$\tan\theta=\frac{x}{r\cos\theta}$
. Equating we get$2x=\frac{x}{r\cos\theta}$
and so$r=\frac{1}{2\cos\theta}$
.Alternatively, we could also have used the fact that the product between the slopes of the tangent and the normal is
$1$
. The slope of the normal in our case is minus the tangent of the angle opposite$\theta$
(can you explain why?), i.e.$\frac{r\cos\theta}{x}$
. Hence$r$
is extracted from$\frac{r\cos\theta}{x}\cdot2x=1$
.