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joe's questions at Yahoo! Answers regarding parametric equations

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MarkFL

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Feb 24, 2012
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Here are the questions:

Mathematics extension 1 (help with 2 parametric questions)?

P(2ap,ap^2) and Q(2aq,aq^2) are two points on the parabola x^2 = 4ay

a) Show that the chord PQ has the equation (p + q) x - 2y = 2apq
b) If P and Q move on the parabola such that pq = 1, where p is not = 0 and q is not = 0, show that the chord PQ (produced) always passes through a fixed point R on the y axis

Question 2

The normal to the parabola x^2 = 2ay at the point T(2at,at^2) cuts the x-axis at X and the y-axis at Y.

a) show the normal at T has equation x + ty = 2at + at^3
b) Show that TX/TY = t^2/2

also u dont need to do both if u cant be bothered but please show all working out
I have posted a link there to this topic so the OP can see my work.
 
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MarkFL

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Feb 24, 2012
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Hello joe,

1.) We are given the points $\left(2ap,ap^2 \right),\,\left(2aq,aq^2 \right)$

a) To determine the line through which the chord $\overline{PQ}$ passes, we first need to determine the slope of the line:

\(\displaystyle m=\frac{\Delta y}{\Delta x}=\frac{aq^2-ap^2}{2aq-2ap}=\frac{q^2-p^2}{2(q-p)}=\frac{q+p}{2}\)

We now have the slope, and we can use either given point (I'll choose point $P$) in the point slope formula:

\(\displaystyle y-ap^2=\frac{q+p}{2}(x-2ap)\)

Multiply through by 2:

\(\displaystyle 2y-2ap^2=(q+p)(x-2ap)\)

Distribute on the right:

\(\displaystyle 2y-2ap^2=(p+q)x-2apq-2ap^2\)

Add $2ap^2$ to both sides:

\(\displaystyle 2y=(p+q)x-2apq\)

Rearrange:

\(\displaystyle (p+q)x-2y=2apq\)

b) Using $pq=1$ and $x=0$, we find the $y$-intercept of $\overline{PQ}$ is found with:

\(\displaystyle -2y=2a\implies y=-a\)

Hence, the $y$-intercept is the fixed point $(0,-a)$.

2.) I am guessing the parabola should be $x^2=4ay$, otherwise the given point is not on the parabola.

a) The slope of the normal line at the point $\left(2at,at^2 \right)$ can be found using a pre-calculus technique or by differentiation.

i) Analysis of the discriminant:

Let's let the tangent line be $y=mx+b$. Substituting into the parabola, we have:

\(\displaystyle x^2=4a(mx+b)\)

\(\displaystyle x^2-4amx-4ab=0\)

In order for the line to be tangent, we require the discriminant to be zero, so we require:

\(\displaystyle (-4am)^2-4(1)(-4ab)=0\)

\(\displaystyle 16a^2m^2+16ab=0\)

\(\displaystyle am^2+b=0\)

Since the given point must pass through point $T$, we also require:

\(\displaystyle at^2=m(2at)+b\implies b=at^2-2amt=at(t-2m)\)

And so we have:

\(\displaystyle am^2+at(t-2m)=0\)

\(\displaystyle m^2+t(t-2m)=0\)

\(\displaystyle m^2-2tm+t^2=0\)

\(\displaystyle (m-t)^2=0\)

\(\displaystyle m=t\)

Thus, the slope of the normal line must be:

\(\displaystyle m=-\frac{1}{t}\)

ii) Differentiation:

The slope of the normal line can be found using \(\displaystyle -\frac{dx}{dy}\).

Begin with the parabola:

\(\displaystyle x^2=4ay\)

Implicitly differentiate with respect to $y$:

\(\displaystyle 2x\frac{dx}{dy}=4a\)

\(\displaystyle -\frac{dx}{dy}=-\frac{2a}{x}\)

Hence:

\(\displaystyle m=\left.-\frac{dx}{dy}\right|_{\left(2at,at^2 \right)}=-\frac{2a}{2at}=-\frac{1}{t}\)

Okay, we now have the slope, and a point, so the normal line is found using the point-slope formula:

\(\displaystyle y-at^2=-\frac{1}{t}(x-2at)\)

Multiply through by $t$:

\(\displaystyle ty-at^3=2at-x\)

\(\displaystyle x+ty=2at+at^3\)

b) $X$ is found by equating $y$ to zero and solving for $x$:

\(\displaystyle x=2at+at^3\)

and so we find:

\(\displaystyle X=\left(2at+at^3,0 \right)\)

$Y$ is found by equating $x$ to zero and solving for $y$:

\(\displaystyle ty=2at+at^3\)

\(\displaystyle y=2a+at^2\)

and so we find:

\(\displaystyle Y=\left(0,2a+at^2 \right)\)

Hence, using the distance formula, we obtain:

\(\displaystyle \overline{TX}=\sqrt{\left(2at-2at-at^3 \right)^2+\left(at^2-0 \right)^2}=\sqrt{\left(-at^3 \right)^2+\left(at^2 \right)^2}=at^2\sqrt{t^2+1}\)

\(\displaystyle \overline{TY}=\sqrt{\left(2at-0\right)^2+\left(at^2-2a-at^2 \right)^2}=\sqrt{\left(2at\right)^2+\left(2a \right)^2}=2a\sqrt{t^2+1}\)

And so we find:

\(\displaystyle \frac{\overline{TX}}{\overline{TY}}=\frac{at^2 \sqrt{t^2+1}}{2a\sqrt{t^2+1}}=\frac{t^2}{2}\)