Question

Let the circle, \((x-\mathrm{a})^{2}+(y-\mathrm{b})^{2}=8\) and the parabola, \(y^{2}=4 x\) have a common tangent at the point \(\mathrm{P}(1,2)\). If \(\mathrm{C}\) is the centre of the circle and \(\mathrm{S}\) is the focus of the parabola, then the area (in sq. units) of \(\triangle \mathrm{CPS}\) is equal to ________.

Hint:

The radius CP is perpendicular to the common tangent at P; use the tangent of the parabola at (1, 2) and the radius 2√2 to locate C.

Answer 1

Correct Answer: 2

Tangent to y² = 4x at (1, 2): 2y = 2(x + 1), i.e. y = x + 1, slope 1.
The circle has radius √8 = 2√2 and touches this line at P, so CP is perpendicular to the tangent: C = (1 + t, 2 - t) with |CP| = |t|√2 = 2√2, so t = ±2.
C = (3, 0) or C = (-1, 4). The focus of the parabola is S = (1, 0).
For C = (3, 0): area = (1/2) |SC| × (height of P above the x-axis) = (1/2)(2)(2) = 2.
For C = (-1, 4): area = (1/2)|(P - S) × (C - S)| = (1/2)|0 × 4 - 2 × (-2)| = 2.
Either way the area is 2 sq. units.
Answer: 2