Cambridge Prelim MX1 Textbook Marathon/Q&A (5 Viewers)

DatAtarLyfe

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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

I take it to mean: x^2 = (28/5)y

In that case simply express in the form: x2 = 4ay

For this question: x^2 = 4* (7/5)*y

so, the focal length "a" is 7/5.
Lol, i was right but deleted my post cause i was like "wait wtf, it's 28/5y"
 

appleibeats

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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

i though the focus length was the distance from the vertex and the focus.

I got the focus was (0,7/5)

and the vertex ( 14/5, 7/5)

so I got the focus length = 14/5

Where did i go wrong?
 

VBN2470

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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

For , vertex is , focus is , hence focal length is .
 

DatAtarLyfe

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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

i though the focus length was the distance from the vertex and the focus.

I got the focus was (0,7/5)

and the vertex ( 14/5, 7/5)

so I got the focus length = 14/5

Where did i go wrong?
Your vertex is incorrect. If you look at your original equation, x^2=28/5y, your vertex is actually (0,0)
To determine the vertex from your equation, you use the standard equation (x-h)^2=4a(y-k), where (h,k) is your vertex. So in your equation, h=0, k=0 and 4a=28/5
Hence your vertex is (0,0), your focal length is 7/5 and thus your focus is (0,7/5)
 

Drongoski

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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

i though the focus length was the distance from the vertex and the focus.

I got the focus was (0,7/5)

and the vertex ( 14/5, 7/5)

so I got the focus length = 14/5

Where did i go wrong?
The focal length is indeed the distance from the vertex to the focus.
 

DatAtarLyfe

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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

i though the focus length was the distance from the vertex and the focus.

I got the focus was (0,7/5)

and the vertex ( 14/5, 7/5)

so I got the focus length = 14/5


Where did i go wrong?
Also, when you have an equation where it is x^2, then the focus is above/below (depending on concavity) the vertex in the Y-DIRECTION. Therefore, the x-oordinate of the vertex and focus will be the same, only the y-oordinates will differ by the magnitude of the focal length.
 

appleibeats

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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

For y = x^2/7

I know it is a cusp. But why does the derivative have a discontinuity at x=0 instead of a zero

y' = 2/7 x^-5/7

doesn't the derivative equal zero when x = 0
 

Speed6

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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

For y = x^2/7

I know it is a cusp. But why does the derivative have a discontinuity at x=0 instead of a zero

y' = 2/7 x^-5/7

doesn't the derivative equal zero when x = 0
That is a silly mistake that can be avoidable.
 

appleibeats

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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

For y = (1 - x ) ^1/4 - 2

the domain is x less than or equal to 1

but y' = 1 / 4(1 - x) ^3/4

so at x = 1

the derivative is undefined. What is the significance of this and do you include the point in the graph?
 

VBN2470

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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

For , y(1) = -2 so you certainly include the point (1,-2) when sketching y = f(x). Derivative is undefined at x = 1 (hence not differentiable at that point), so the graph of y = f(x) at x = 1 will contain some sort of 'cusp' or 'sharp' point (i.e. the slope of the graph of f is undefined). You can check this graphically.
 
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DatAtarLyfe

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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

For , y(1) = 0 so you certainly include the point (1,0) when sketching y = f(x). Derivative is undefined at x = 1 (hence not differentiable at that point), so the graph of y = f(x) at x = 1 will contain some sort of 'cusp' or 'sharp' point (i.e. the slope of the graph of f is undefined). You can check this graphically.
This is the critical point, right?
 

appleibeats

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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Using dy/dx = dy/dt / dx/dt , find any stationary points and sketch the graphs of the functions:

a) x = 6t , y = 3t^2

dy/dx = 6t /6 = t

so dy/dx = 0 when t = 0

Is that right and if so how do I go about graphing the function now?
 

VBN2470

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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

So you found that when t = 0 , which means the derivative is 0 at the point . To sketch the graph of the curve defined by those parametric equations, we eliminate the parameter t to get:
Or you could square x first and it will make it bit easier maybe. You will end up having to sketch the standard parabola .
 

appleibeats

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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

A parabola has Cartesian equation x^2 = - 8/3 y

I have found that the focal length = -2/3

How doe I now find the parametric equations of the parabola?

And hence find the equation of the chord PQ given that the point P has parameter 2 and the point Q has parameter -1. ( Assuming the equation of a chord)

I assume using: y = ( p + q / 2)x - apq
 

VBN2470

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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Remember, focal length is always positive, so it should be , since your parabola is of the form (concave down). Parametric equation of the parabola will given by , where a is a positive constant and t is your parameter. So your parametric equations will be . The endpoints of the chord will be given by (substituting t = 2 and t = -1 respectively). Then, you can find the Cartesian equation of the chord using standard coordinate geometry. Equation of the chord should be 3x + 6y + 8 = 0.
 
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appleibeats

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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

A parabola has parametric equations x = -8t - 7 , y = -4t^2 + 2

i) state the focal length
ii) Without finding the Cartesian equation, write down the coordinates of the vertex, and the equation of the directrix

I began by eliminating the parameter t :

t = x + 7 / -8

y = -4( x + 7 / -8 )^2 + 2

But isn't that find the Cartesian equation? Is there an easier way?
 

InteGrand

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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

A parabola has parametric equations x = -8t - 7 , y = -4t^2 + 2

i) state the focal length
ii) Without finding the Cartesian equation, write down the coordinates of the vertex, and the equation of the directrix

I began by eliminating the parameter t :

t = x + 7 / -8

y = -4( x + 7 / -8 )^2 + 2

But isn't that find the Cartesian equation? Is there an easier way?


 

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