CSSA 2004 question (1 Viewer)

[eskimoh]

http://img72.imageshack.us/my.php?image=mathsqx6.jpg



could someone post their answers to this question, cos i dont have answers to this paper and i want to know whether i was correct


thnx~

 

[eskimoh]

bumpppppppp

 

[trailblazer]

i) y=ax^2 and y= 12x + 3

so by equating y, you will get

ax^2 = 12x + 3 then moving to one side, you will get a quadratic equation in terms of x and a

thus you get, ax^2 - 12x- 3 = 0

ii) Now they tell you that the line is a tangent, therefore it touches the parabola at one point. This means there is a double root, i.e one solution. From this, we can say that the discriminant = 0 (remember yr10 quadratics)

So, in the equation ax^2 - 12x- 3 = 0,
discriminant= b^2 - 4ac= 144 - 4 . -3 . a
= 144 + 12a

But discriminant also equals to 0, so

0 = 144 + 12a
-144 = 12a
a=-12

So, the parabola's equation is y=-12x^2

iii) ( -1/2 , -3 )