Cambridge Prelim MX1 Textbook Marathon/Q&A (1 Viewer)

appleibeats · Nov 3, 2015

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

Have a line that intesects the coordinate axes at ( 0 , 2a + ap^2) and (2ap + ap^3 , 0 )

Now find the area of the triangle formed by the two intercepts and the origin.

So using 1/2 . length . width

I get

A = 1/2 a^2 ( 2 + p^2) (2 + p^3)

Not sure where to go from here.

Answer is 1/2 a^2 |p| (p^2 + 2)^2

Not sure why there is an absolute value p??

InteGrand · Nov 3, 2015

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

appleibeats said:
Have a line that intesects the coordinate axes at ( 0 , 2a + ap^2) and (2ap + ap^3 , 0 )

Now find the area of the triangle formed by the two intercepts and the origin.

So using 1/2 . length . width

I get

A = 1/2 a^2 ( 2 + p^2) (2 + p^3)

Not sure where to go from here.

Answer is 1/2 a^2 |p| (p^2 + 2)^2

Not sure why there is an absolute value p??

$The area $\mathcal{A}$ is given by$

$\begin{align*}\mathcal{A} &= \left |\frac{1}{2}\times \left(2a + ap^2 \right)\times \left( 2ap + ap^3 \right) \right| \\ &=\left| \frac{1}{2}a^2 \left(2+p^2 \right)\left(2p+p^3\right)\right| \\ &= \frac{1}{2}a^2 \left(2+p^2 \right)\left|p \left(2+p^2 \right)\right| \\ &= \frac{1}{2}a^2 |p|\left (2+p^2 \right)\left (2+p^2\right).\end{align*}$

$We need to take absolute values because those intercepts could be negative ($p$ could be negative), and areas are positive. Also note that things like $\left| 2+p^2\right|$ are equal to $2+p^2$ since they are positive, so we can just take off absolute values from them.$

appleibeats · Nov 3, 2015

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

Find the parameters of the points of contact of the tangents to x = 10t , y = 5t^2 from the point P (24,-5)

Not sure how to approach this question.

InteGrand · Nov 3, 2015

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

appleibeats said:
Have a line that intesects the coordinate axes at ( 0 , 2a + ap^2) and (2ap + ap^3 , 0 )

Now find the area of the triangle formed by the two intercepts and the origin.

So using 1/2 . length . width

I get

A = 1/2 a^2 ( 2 + p^2) (2 + p^3)

Not sure where to go from here.

Answer is 1/2 a^2 |p| (p^2 + 2)^2

Not sure why there is an absolute value p??

$So firstly, you left out a $p$ in your last factor \Big{(}should have been a $\left( 2p + p^3\right)$\Big{)}. And secondly, the formula to use should have been $\mathcal{A} = \frac{1}{2}\left| 2a + ap^2 \right| \left|2ap + ap^3 \right|$ rather than $\mathcal{A} = \frac{1}{2}\left( 2a + ap^2 \right) \left(2ap + ap^3 \right)$, since we're not told that the intercepts have the same sign (your formula would have given the correct expression for the area if and only if the two intercepts were either both positive or both negative, so that their product came out to be positive, like an area should be; but we can't assume this as we're not told it).$

InteGrand · Nov 3, 2015

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

appleibeats said:
Find the parameters of the points of contact of the tangents to x = 10t , y = 5t^2 from the point P (24,-5)

Not sure how to approach this question.

$Let the desired points be $X=\left(10t,5t^2\right)$ (we need to find the values of $t$, and you should end up with two values for it). We know that the slope of the tangent at $X$ is $t$ (easy to prove; you should know how to do this). Also, the slope of the tangent is the slope of $PX$, which is $m_{PX}=\frac{5t^2 -(-5)}{10t - 24}$ (slope formula). So we can equate this expression to $t$, and rearrange to get a quadratic in $t$, which you can easily solve, and the solutions are the answers.$

appleibeats · Nov 4, 2015

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

Use the discriminant to show that mx - y - m^2 = 0 touches the parabola x^2 = -4y for all values of m

Hence find the equations of the tangents to x^2 = -4y through the point A (1,2)

appleibeats · Nov 8, 2015

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

How to solve for z:

z^2 - z + ( 1 + i) = 0

Using The Quadratic method with finding out lander .

Really confused on what to do.

InteGrand · Nov 8, 2015

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

appleibeats said:
How to solve for z:

z^2 - z + ( 1 + i) = 0

Using The Quadratic method with finding out lander .

Really confused on what to do.

$Since $(1+i)$ is a constant, we just apply the quadratic formula as usual, with $a=1$, $b=-1$, and $c=1+i$.$

$You will need to find the square root of a complex number to simplify the answer. Have you learnt how to do this yet?$

crimp · Nov 10, 2015

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

Cambridge 3U 9G
Could someone please help me out with part b) and d)?
12. Tangents are drawn to the parabola y=x^2 from the point T(1,-1). These tangents touch the parabola at P and Q.
a) Obtain a quadratic equation whose roots are the x-coordinates of P and Q and write down the sum and product of these roots
b) Find a quadratic equation whose roots are the y-coordinates of P and Q and write down the sum and product of these roots
c) Prove this identity (p-q)^2=(p+q)^2-4pq
d) Use the distance formula and this identity to find the length of the chord PQ

Thanks!!

Drongoski · Nov 10, 2015

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

$x^2 = y = 4 \times \frac {1}{4} \times y \implies a = \frac {1}{4} \\ \\ $Eqn of chord of contact is: $ xx_0 = 2a(y + y_0) \implies x \times 1 = 2 \times \frac {1}{4} (y - 1)\\ \\ $i.e. $y = 2x + 1\\ \\ $Where this chord of contact intersect the parabola: $ y = x^2 = 2x + 1\\ \\ \therefore x^2 - 2x + 1 = 0 $ whose 2 roots $ x_1 = 2ap $ and $ x_2 = 2aq $ satisfy: $\\ \\ $sum of roots $2ap + 2aq = -(-2) = 2 $ and product of roots $ ap^2 \times aq^2 = -1\\ \\ $The parabola and the chord of contact, where they intersect, in terms of y instead of x give us: $\\ \\ x^2 = y = (\frac {1}{2} (y-1))^2 \implies y^2 - 6y + 1 = 0 $ with roots satisfying: $ \\ \\ $ sum of roots $= - (-6) = 6 $ and product of roots $ = 1\\ \\ \\$(c) The identity $ \\ \\$Now :$ (p-q)^2 = p^2 -2pq + q^2 = p^2 + 2pq +q^2 - 4pq = (p+q)^2 - 4pq\\ \\ $(d)$ \\ \\$ $$With $ P(2ap,ap^2) = (\frac {1}{2}p. \frac{1}{4} p^2)$ and $ Q(2aq,aq^2) = (\frac {1}{2}q, \frac {1}{4}q^2)\\ \\ PQ^2 = (\frac {1}{2}p - \frac{1}{2}q)^2 + (\frac{1}{4}p^2 - \frac {1}{4}q^2)^2 = (\frac {1}{4})^2(p-q)^2 (4+(p+q)^2)\\ \\ = (\frac{1}{4})^2((p+q)^2 - 4pq)\times (4 + (p+q)^2)\\ \\ = (\frac {1}{4})^2 (4^2 - 4 \times (-4)) \times (4+ 4^2) = 40 \\ \\ \therefore PQ = 2 \sqrt {10}$

crimp · Nov 10, 2015

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

Drongoski said:
$x^2 = y = 4 \times \frac {1}{4} \implies a = \frac {1}{4} \times y\\ \\ $Eqn of chord of contact is: $ xx_0 = 2a(y + y_0) \implies x \times 1 = 2 \times \frac {1}{4} (y - 1)\\ \\ $i.e. $y = 2x + 1\\ \\ $Where this chord of contact intersect the parabola: $ y = x^2 = 2x + 1\\ \\ \therefore x^2 - 2x + 1 = 0 $ whose 2 roots $ x_1 = 2ap $ and $ x_2 = 2aq $ satisfy: $\\ \\ $sum of roots $2ap + 2aq = -(-2) = 2 $ and product of roots $ ap^2 \times aq^2 = -1\\ \\ $The parabola and the chord of contact, where they intersect, in terms of y instead of x gives us: $\\ \\ x^2 = y = (\frac {1}{2} (y-1))^2 \implies y^2 - 6y + 1 = 0 $ with roots satisfying: $ \\ \\ $ sum of roots $= - (-6) = 6 $ and product of roots $ = 1\\ \\ \\$(c) The identity $ \\ \\$Now :$ (p-q)^2 = p^2 -2pq + q^2 = p^2 + 2pq +q^2 - 4pq = (p+q)^2 - 4pq\\ \\ $(d)$ \\ \\$ $$With $ P(2ap,ap^2) = (\frac {1}{2}p. \frac{1}{4} p^2)$ and $ Q(2aq,aq^2) = (\frac {1}{2}q, \frac {1}{4}q^2)\\ \\ PQ^2 = (\frac {1}{2}p - \frac{1}{2}q)^2 + (\frac{1}{4}p^2 - \frac {1}{4}q^2)^2 = (\frac {1}{4})^2(p-q)^2 (4+(p+q)^2)$

Thanks man!

appleibeats · Nov 10, 2015

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

P (2ap, ap^2) and Q (2aq, aq^2) are variable point on the parabola x^2 = 4ay. R is the intersection of the tangent at P and the line through Q parallel to the axis of the parabola, while U is the intersection of the tangent at Q and the line through P parallel to the axis.

a) Show that PQRU is a parallelogram

b) if p > q, show that parallelogram PQRU has are 2a^2(p - q)^3 units^2

I have done part a,

Not sure how to approach part b.

InteGrand · Nov 10, 2015

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

appleibeats said:
P (2ap, ap^2) and Q (2aq, aq^2) are variable point on the parabola x^2 = 4ay. R is the intersection of the tangent at P and the line through Q parallel to the axis of the parabola, while U is the intersection of the tangent at Q and the line through P parallel to the axis.

a) Show that PQRU is a parallelogram

b) if p > q, show that parallelogram PQRU has are 2a^2(p - q)^3 units^2

I have done part a,

Not sure how to approach part b.

A way to find the area of a parallelogram is the find the perpendicular distance from one vertex to the line through the side opposite, and multiply this by the length of that line (i.e. http://www.math.fsu.edu/~wooland/hm2ed/Part3Module8/Sol2/graphics/Sol22.jpeg ).

(Alternatively, we can use the absolute value of the determinant of a 2×2 matrix.)

appleibeats · Nov 12, 2015

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

Find and describe the locus of points in the Argand diagram which correspond to zz(conjugate) = (Re( z - 1 + 3i))^2

LHS is z times z (conjugate)

Answer is a parabola with focus at the origin and the directrix at x = 1

InteGrand · Nov 12, 2015

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

appleibeats said:
Find and describe the locus of points in the Argand diagram which correspond to zz(conjugate) = (Re( z - 1 + 3i))^2

LHS is z times z (conjugate)

Answer is a parabola with focus at the origin and the directrix at x = 1

$Let $z=x+iy$, where $x=\Re({z})$ (real part of $z$) and $y=\Im ({z})$ (imaginary part of $z$). Note that $z\bar{z}=|z|^2$. Also, $\Re \left(z-1+3i \right)=x-1$ (since the real part of a sum of complex numbers is the sum of the real parts). So we have $|z|^2 = \left(x-1\right)^2\Rightarrow |z| = \left | x-1\right|$ (absolute value of $x-1$).$

$In other words, $z$ moves so that its distance from the \textbf{origin} (which is $|z|$) is equal to its horizontal distance from the \textbf{line \textit{x} = 1} (which is $|x-1|$). This means (by the geometric definition of a parabola), that the locus is a parabola whose focus is the \textbf{origin} and directrix is the \textbf{line \textit{x} = 1}. This is a parabola that has vertex at $\left(\frac{1}{2},0\right)$ and focal length $\frac{1}{2}$. It is a leftward-opening parabola, so its equation is of the form $\left(y-k\right)^2 = -4a\left(x-h\right)$ We have $h=\frac{1}{2}$ ($x$-value of the vertex), $k=0$ ($y$-value of the vertex), and $a=\frac{1}{2}$ (focal length), so the equation of the locus is $y^2 = -2\left(x-\frac{1}{2}\right)$.$

appleibeats · Nov 13, 2015

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

Find the values p and q given that 1 + 2i is a root of the equation z^3 + pz + q = 0, where p and q are real.

InteGrand · Nov 14, 2015

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

appleibeats said:
Find the values p and q given that 1 + 2i is a root of the equation z^3 + pz + q = 0, where p and q are real.

Since we know 1 + 2i is a root, we can sub. it into that equation, then expand everything and group real and imaginary parts on the LHS, then equate the real and imaginary parts to 0, since the RHS is 0. You'll end up with two equations in two unknowns, which you can solve for p and q.

appleibeats · Nov 14, 2015

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

If 4 - tanE = 5sinEcosE, show that x = tanE is a root of the equation x^3 - 4x^2 + 6x - 4 = 0

InteGrand · Nov 14, 2015

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

appleibeats said:
Find the values p and q given that 1 + 2i is a root of the equation z^3 + pz + q = 0, where p and q are real.

$So basically, we have $\left(1+2i \right)^3 + p\left(1+2i \right) + q = 0$. Therefore,$

$\begin{align*}LHS&= 1 + 3\cdot (2i) + 3 \cdot (2i)^2 + (2i)^3 + p + 2ip + q \quad (\text{expanding})\\ &= 1 + 6i -12 - 8i + p + 2ip +q \quad (\text{as }i^2 = -1\text{ and }i^3 = -i)\\ &=\left(-11 + p + q \right)+ i \left(-2 + 2p \right) \quad (\text{writing in the form }a+ib,\text{ where }a\text{ and }b\text{ are real}).\end{align*}$

$Since $LHS=0 = 0+0i$, we must have both the real and imaginary parts of $LHS$ be 0, so $-2+2p=0\Rightarrow p = 1$ and $-11+p+q = 0 \Rightarrow q = 11-p=11-1 = 10$.$

InteGrand · Nov 14, 2015

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

appleibeats said:
If 4 - tanE = 5sinEcosE, show that x = tanE is a root of the equation x^3 - 4x^2 + 6x - 4 = 0

$Since $\sin E \cos E = \frac{1}{2} \sin 2E$, we have $4 - \tan E = 5 \sin E \cos E \Rightarrow 4 - \tan E = \frac{5}{2} \sin 2E$. Letting $x=\tan E$, we have by the $t$-formulae that $\sin 2E = \frac{2x}{1+x^2}$. Substituting these into the previous equation, we have $4-x = \frac{5}{2}\times \frac{2x}{1+x^2}$$

$\begin{align*}\Rightarrow 4-x &= \frac{5x}{1+x^2} \\ \Rightarrow (4-x) \left(1+x^2\right) &= 5x \\ \Rightarrow 4 + 4x^2 -x - x^3 &= 5x \\ \Rightarrow x^3 -4x^2 +6x-4 &= 0 \quad (\text{rearranging})\end{align*}$

$as required.$

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