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# 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 $$\theta$$.

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## Warm-up Questions

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.

## Hints

• Hint 1
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?
• Hint 2
What is the angle made by the tangent at $$P$$ with the $$x$$-axis in terms of $$\theta$$?
• Hint 3
What is another expression of the slope of the tangent at $$P$$ in terms of this angle.
• Hint 4
Can 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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