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.

The Computer Laboratory

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 \(\theta\).

The above links are provided as is. They are not affiliated with the Climb Foundation unless otherwise specified.

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

If you have queries or suggestions about the content on this page or the CSAT Practice Platform then you can write to us at oi.foo[email protected]. Please do not write to this address regarding general admissions or course queries.