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# Practice Paper 4 Question 4

The parabola $$y=x^2+\frac{1}{4}$$ is eventually intersected at two points by the line $$y=tx,$$ where $$t$$ is the time increasing linearly from $$0$$ (hence the line rotates from a horizontal towards a vertical position). Determine the speed at which the segment linking the two intersection points grows.

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## Hints

• Hint 1
Where do the two curves intersect?
• Hint 2
Find, in terms of $$t,$$ the coordinates of these points.
• Hint 3
... and the length of the segment joining the two points.
• Hint 4
How could you find the speed at which it grows from this length?

## Solution

We first find where the points of intersection are, in terms of $$t.$$ We equate the two curves to get $$x^2-tx+\frac{1}{4} = 0.$$ Solving for $$x$$ gives us their $$x$$-coordinates: $$x_{A,B} = \frac{t\pm\sqrt{t^2-1}}{2}.$$

The two points $$A$$ and $$B$$ lie on the line $$y=tx$$ so we substitute in the above $$x$$-coordinates to get their $$y$$-coordinates. Thus the two points are: $$A = \big(\frac{t-\sqrt{t^2-1}}{2},\frac{t^2-t\sqrt{t^2-1}}{2}\big)$$ and $$B = \big(\frac{t+\sqrt{t^2-1}}{2},\frac{t^2+t\sqrt{t^2-1}}{2}\big)$$

The length of the segment joining $$A$$ and $$B$$ is their Euclidean distance given by $$D_{A,B}(t) = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}.$$ Substituting and simplifying yields $$D_{A,B}(t) = \sqrt{t^4-1}.$$

To get the rate at which this distance increases, we differentiate with respect to $$t$$ to get the result $$D'_{A,B}(t) = \frac{2t^3}{\sqrt{t^4-1}}.$$

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