The CSAT and Practice[+] are designed by the Climb Foundation to help candidates. We are advocates for more opportunity to shine and less opportunity to fail, and we strive to level the playing field. 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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