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

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 $-\frac{r\cos\theta}{x}\cdot2x=-1$.