Rectangle under a parabolic arch

Rectangle under a parabolic arch

Secondary 2
8 marks
TGM Original Questions

Optional extension

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]
  2. Express \(PS\) and \(RS\) in terms of \(x,y\).[2]
  3. Form and solve an equation for the valid \(x\).[2]
  4. Find the area of the rectangle.[2]

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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