# Practice Paper 1 Question 10

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 \(\theta\).

## Related topics

## Warm-up Questions

- What is the slope of the tangent to the curve \(y = 2x^2+3x-2\) 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: \(y=2x\), \(y=2x-2\), \(y=-\frac{1}{2}x+3\). For each pair of lines, state whether they are parallel or perpendicular.

## Hints

- Hint 1Let \(P\) 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?
- Hint 2What is the angle made by the tangent at \(P\) with the \(x\)-axis in terms of \(\theta\)?
- Hint 3What is another expression of the slope of the tangent at \(P\) in terms of this angle.
- Hint 4Can you extract the tangent of this angle from the right triangle with sides \(r,\) \(x\) and \(r\cos(\theta)?\)

## Solution

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

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