# 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.

## Related topics

## Hints

- Hint 1Where do the two curves intersect?
- Hint 2Find, in terms of \(t,\) the coordinates of these points.
- Hint 3... and the length of the segment joining the two points.
- Hint 4How 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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