Rectangle under a parabolic arch

Rectangle under a parabolic arch

Secondary 2
TGM Original Questions

The arch \(y=49-x^2\) meets the axes at \(A,B,C\). Rectangle \(PQRS\) is symmetric about the \(y\)-axis, with \(P=(x,y)\) on the arch and \(PS:RS=20:3\).

  1. Find \(A,B,C\) and the symmetry line.
  2. Express \(PS\) and \(RS\) in terms of \(x,y\).
  3. Form and solve an equation for the valid \(x\).
  4. Find the area of the rectangle.

Solution:

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Answer:(a) \(A=(-7,0),B=(7,0),C=(0,49)\), symmetry \(x=0\); (b) \(PS=y,RS=2x\); (c) \(x=3,y=40\); (d) \(240\) square units.

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