2016 VJC Promo Q6

2016 VJC Promo Q6

7 marks
Promo

The shape of a mirror is in the form of a curve \(C\) given by the equation

\({{y}^{2}}=-2x+4\)

  1. Sketch \(C\), indicating clearly the coordinate of the points where the curve crosses the axes.[2]
  2. Find the equation of the normal to \(C\) at the point \(P\left( -\frac{1}{2}{{k}^{2}}+2,k \right)\), where \(k>0\).[2]

A man with a torchlight is standing at the point \(A\left( a,k \right)\), where \(a<-\frac{1}{2}{{k}^{2}}+2\) , He shines the torchlight at \(P\) and the beam of light is reflected by the mirror . The reflected beam of light intersects the \(x-\) axis at the point \(B\). The normal to \(C\) at \(P\) bisects the angle \(APB\).

  1. Show that the \(x\)- coordinate of \(B\) is independent of \(k\).[3]

Video Solution:

Video Solution

Video solution locked

A free account saves your progress; this video needs tuition enrolment.

Solution:

Solution locked

680+ questions here come with free worked solutions. A free account saves your questions and progress; this one needs tuition enrolment.

Create a free account to practise similar questions.

Answer:(ii) \(y = kx + \dfrac{1}{2}k^{3} - k\)

Need help? Join our JC Math tuition classes.

Learn more