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Tw!stedGemz

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i have this question for a maths assignment due in on moday that i have no idae how to do, if anyone can tell me how or just tell me where to start it would be greatly appreciated.

A parabola has its equation in the form y=Ax (the x is squared)
Where A is a constant. The line y=10x+10 is a tangent ot the parabola
y=ax (squared).
a) find the value of A
b) Sketch the parabola and the tangent line, showing the coordinates of the point of contact.
c) Find the coordinates of the focus and the equation of the directrix of the parabola


... i can probably do b) and c) myself but i cant figure out a) and i need that to do them other two XD
 

ssglain

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A bit late for you maths assignment, but just in case you still wanted to know the solution:

Let's find the points of intersection of y = 10x + 10 and y = Ax² by solving the two simultaneously:
Ax² = 10x + 10
Ax² - 10x - 10 = 0

Now, keep in mind that for the line y = 10x + 10 to be a tangent to the parabola y = Ax² there must exist only one point of contact. So we require for this quadratic equation to have only one solution. That is, we require discriminant = 0

Discriminant = (-10)² - 4A(-10) = 0
100 + 40A = 0 --> .: A = -5/2

Hence the equation of the parabola is y = (-5/2)x²

b) and c) you should be able to do from there.
 

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