Cambridge Prelim MX1 Textbook Marathon/Q&A (3 Viewers)

appleibeats · Jul 11, 2015

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

I don't understand how you determine that the parabola is downward opening, and how a = - 4.

InteGrand · Jul 11, 2015

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

appleibeats said:
I don't understand how you determine that the parabola is downward opening, and how a = - 4.

$Since $x-(-7)=-8t=2at$, $2a=-8\Rightarrow a = -4$ (you can also check this with $y-2=-4t^2 = at^2$, which implies that $a=-4$). The parabola is downward facing because $a<0$. (If $a>0$, we would have had the parabola be upward facing.)$

appleibeats · Jul 11, 2015

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

Consider the parabola x^2 = 6y

i) Show that the point P (-2, 2/3) lies on the parabola
ii) Find the coordinates of Q, given that the chord PQ passes through the point (1,1)

For part i) ,

I subbed in x in the LHS and y in the RHS and they equal each other, show thus it lies on the parabola.

Not sure how to answer ii).

InteGrand · Jul 11, 2015

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

appleibeats said:
Consider the parabola x^2 = 6y

i) Show that the point P (-2, 2/3) lies on the parabola
ii) Find the coordinates of Q, given that the chord PQ passes through the point (1,1)

For part i) ,

I subbed in x in the LHS and y in the RHS and they equal each other, show thus it lies on the parabola.

Not sure how to answer ii).

For (ii), we can find the equation of the line PQ since you know two points on it (find the slope using gradient formula, then use point-slope form). Once we have this equation, we just need to solve simultaneously with the parabola's equation to find the intersection points (one of these points will just be P; the other solution we get will be Q's coordinates).

VBN2470 · Jul 11, 2015

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

For part (i), you simply substitute in the values and show that the LHS and RHS are equal.
For part (ii), find the Cartesian equation of the chord, using the points $P(-2, \frac{2}{3})$ and $(1,1)$ you should find that the equation of the chord is x - 9y + 8 = 0. Then sub. in $y=\frac{x^2}{6}$ into the equation of the chord to find the points of intersection between the chord and parabola. Point Q should be $Q(\frac{8}{3}, \frac{32}{27})$ , since your other point will just be P.

appleibeats · Jul 11, 2015

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

The point P (4,3) and Q(2aq, aq^2) lie on the parabola x^2 = 4ay

i) Find the focal length, a
ii) Find the equation of the chord PQ
iii) Hence, find the coordinates of Q, given that PQ is a focal chord

For i) do you just sub in the points of P and so you get a =4/3

ii) Now do you just leave Q as (8/3 q, 4/3 q^2) and use point gradient formula?

iii) does this have something to do with pq = -1 ?

VBN2470 · Jul 11, 2015

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

For (i) you are correct, $a=\frac{4}{3}$ .
For (ii), leave Q as $Q(\frac{8}{3}q, \frac{4}{3}q^2)$ and use point-gradient formula to find the equation of the chord.
For (iii), since PQ is a focal chord, it passes through the focus of the parabola $S(0, \frac{4}{3})$ , sub. this into the equation of the chord to find the value of q, and sub. this into your point Q to find the coordinates of Q.

rand_althor · Jul 11, 2015

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

appleibeats said:
The point P (4,3) and Q(2aq, aq^2) lie on the parabola x^2 = 4ay

i) Find the focal length, a
ii) Find the equation of the chord PQ
iii) Hence, find the coordinates of Q, given that PQ is a focal chord

For i) do you just sub in the points of P and so you get a =4/3

ii) Now do you just leave Q as (8/3 q, 4/3 q^2) and use point gradient formula?

iii) does this have something to do with pq = -1 ?

i and ii) Yes.
iii) You'll have to sub in the coordinates of the focus into the equation of the chord.

appleibeats · Jul 11, 2015

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

I subbed in ( 0, 4/3) into the equation:

and I eventually got the quadratic:

(2q -3) (3q + 2) = 0

I am not sure which value of q i am meant to take. Which value can I disregard?

VBN2470 · Jul 11, 2015

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

When $q=\frac{3}{2}$ , sub. it into Q to find that the coordinates of Q are $Q(4,3)$ (which is the same as P, hence you disregard this). The solution $q=-\frac{2}{3}$ will give the coordinates of Q to be $Q(-\frac{16}{9}, \frac{16}{27})$ which should be the correct answer.

appleibeats · Jul 11, 2015

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

A graph has parametric equations x = ct, y = c/t , where c is a constant

i) Find the Cartesian equation of the graph
ii) The points A (at, a/t) and B (bt, b/t) lie on the graph. Find the gradient of AB.
iii) Hence, find an expression for the equation of the chord AB.

i) c = x/t
y = x/t / t = x / t^2

ii) Gradient = b/t - a/t / bt - at
= b - a / t^2(b - a)
= 1/t^2

Is that correct?

iii) Then using the point gradient formula,

I get y = x / t^2

Is that right and why does it equal the Cartesian equation of the graph found in part i)?

VBN2470 · Jul 11, 2015

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

c is a constant and t is a parameter, you've mixed the two up. Instead, after eliminating your parameter t, you should have the Cartesian equation $xy=c^2$ . For parts (ii) and (iii), you are on the right track but try to see how you can eliminate t in your expressions and work from there.

appleibeats · Jul 11, 2015

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

So now I have

xy = c^2 as the cartesian eqn of the graph

So does that mean for ii), since t = x/c

Its just: sub t = x/c into 1/t^2

so Gradient of AB = c^2 / x^2

And the Equation of the chord AB is y = c^2/x

InteGrand · Jul 12, 2015

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

appleibeats said:
So now I have

xy = c^2 as the cartesian eqn of the graph

So does that mean for ii), since t = x/c

Its just: sub t = x/c into 1/t^2

so Gradient of AB = c^2 / x^2

And the Equation of the chord AB is y = c^2/x

Your original method and answer for the gradient was correct (gradient is 1/t^2).

Then use point-gradient form: equation of line is y - at = (1/t^2)•(x - a/t), which simplifies to your original answer of y = (1/t^2)•x.

Incidentally, the only way the distinct points A and B can lie on the hyperbola are if $(a,b) = (c,-c)$ or $(a,b) = (-c,c)$ .

appleibeats · Jul 12, 2015

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

For Q 17 from 7D.

Show that the line x + y + 2 = 0 is a tangent to y = x^3 - 4x and find the point of contact. (Hint: find the equations of the tangents parallel to x + y + 2 = 0 , and show that one of them is this very line)

I am very confused. I thought I should first find the point of contact and once doing so If the gradient of the line and cubic are the same at such point then they are tangent. Is that right?

To find point of contact I subbed in y into the line:

And get: x^3 - 3x + 2 = 0

Not sure where to go now.

leehuan · Jul 12, 2015

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

This is one way of doing it using polynomials:

$$If the line is tangent, then when you solve simultaneously one of the roots is at least a DOUBLE root.$ \\ $Let $ f\left( x \right) ={ x }^{ 3 }-3x+2\\ f^{ \prime }\left( x \right) =3{ x }^{ 2 }-3 \\ $If a double root exists, the root is shared between $ f\left( x \right)$ and $ f^{ \prime }\left( x \right) \\ $Solving $f^{ \prime }\left( x \right)=0$ gives $x=\pm 1 \\ $Then we find that $f\left( 1 \right) =0$ and hence a double root occurs at $x=1$

$\\ $Thus going back to the original question, we find that the line IS tangent, and we have found the point where it is tangential as well. So we can just sub back into $x+y+2=0$ to find the coordinates.$

But of course, I didn't use the hint. Hence it was most likely not the desired method.

VBN2470 · Jul 12, 2015

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

$x^3-3x+2=0 \Rightarrow (x-1)^2(x+2)=0$ , so your points of intersection will be $(-2,0)$ and $(1,-3)$ . Also, since x = 1 is a double root, it follows that it the line x + y + 2 = 0 must be tangential to the cubic. Not sure what the hint wants us to do, unless they want us to sketch it and show that it is tangential graphically

leehuan · Jul 12, 2015

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

VBN2470 said:
$x^3-3x+2=0 \Rightarrow (x-1)^2(x+2)=0$ , so your points of intersection will be $(-2,0)$ and $(1,-3)$ . Also, since x = 1 is a double root, it follows that it the line x + y + 2 = 0 must be tangential to the cubic. Not sure what the hint wants us to do, unless they want us to sketch it and show that it is tangential graphically

Factorising the cubic equation generally requires the 3U student to test factors of the constant term before performing long division or division by inspection. Although I probably would've done it that way, there is a problem in that we don't know if all roots are rational... at least, not at the start.

appleibeats · Jul 12, 2015

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

leehuan said:
This is one way of doing it using polynomials:

$$If the line is tangent, then when you solve simultaneously one of the roots is at least a DOUBLE root.$ \\ $Let $ f\left( x \right) ={ x }^{ 3 }-3x+2\\ f^{ \prime }\left( x \right) =3{ x }^{ 2 }-3 \\ $If a double root exists, the root is shared between $ f\left( x \right)$ and $ f^{ \prime }\left( x \right) \\ $Solving $f^{ \prime }\left( x \right)=0$ gives $x=\pm 1 \\ $Then we find that $f\left( 1 \right) =0$ and hence a double root occurs at $x=1$

$\\ $Thus going back to the original question, we find that the line IS tangent, and we have found the point where it is tangential as well. So we can just sub back into $x+y+2=0$ to find the coordinates.$

But of course, I didn't use the hint. Hence it was most likely not the desired method.

Now subbing x = 1, you get y = -3 and so point of contact is (1, -3)

But answer in the book also says at ( -1, 3) the tangent is x + y -2 = 0

Not sure what it means?

appleibeats · Jul 12, 2015

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

How do you answer q 19)

Find the points where the line x + 2y = 4 cuts the parabola y = ( x -1)^2, and show that the line is the normal to the curve at one of these points.

I made x = 4 - 2y

and subbed it in to the parabola and got:

4y^2 - 13y + 9 = 0

(4y - 9) (y -1) = 0

Therefore points are ( 2,1) and the second point I am not sure: when I sub y = 9/4 into the parabola I get x = 5/2 but when into the line x = 8/9

Not sure why I am getting different results. And how to show that at one of these points the line is the normal to the curve.

Cambridge Prelim MX1 Textbook Marathon/Q&A (3 Viewers)

Member

Well-Known Member

Member

Well-Known Member

Well-Known Member

Member

Well-Known Member

Active Member

Member

Well-Known Member

Member

Well-Known Member

Member

Well-Known Member

Member

Well-Known Member

Well-Known Member

Well-Known Member

Member

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 3)