MHB Joe's questions at Yahoo Answers regarding parametric equations

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The discussion revolves around solving parametric equations related to a parabola defined by x^2 = 4ay. The first question involves proving the equation of the chord PQ and demonstrating that it passes through a fixed point on the y-axis when pq = 1. The second question focuses on finding the equation of the normal at a point on the parabola and establishing a relationship between the distances TX and TY. The solutions provided include detailed calculations and derivations for both parts of each question. The discussion emphasizes the importance of showing all working out for clarity and understanding.
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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 don't need to do both if u can't 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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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:

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

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

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

Multiply through by 2:

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

Distribute on the right:

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

Add $2ap^2$ to both sides:

$$2y=(p+q)x-2apq$$

Rearrange:

$$(p+q)x-2y=2apq$$

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

$$-2y=2a\implies y=-a$$

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:

$$x^2=4a(mx+b)$$

$$x^2-4amx-4ab=0$$

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

$$(-4am)^2-4(1)(-4ab)=0$$

$$16a^2m^2+16ab=0$$

$$am^2+b=0$$

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

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

And so we have:

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

$$m^2+t(t-2m)=0$$

$$m^2-2tm+t^2=0$$

$$(m-t)^2=0$$

$$m=t$$

Thus, the slope of the normal line must be:

$$m=-\frac{1}{t}$$

ii) Differentiation:

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

Begin with the parabola:

$$x^2=4ay$$

Implicitly differentiate with respect to $y$:

$$2x\frac{dx}{dy}=4a$$

$$-\frac{dx}{dy}=-\frac{2a}{x}$$

Hence:

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

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

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

Multiply through by $t$:

$$ty-at^3=2at-x$$

$$x+ty=2at+at^3$$

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

$$x=2at+at^3$$

and so we find:

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

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

$$ty=2at+at^3$$

$$y=2a+at^2$$

and so we find:

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

Hence, using the distance formula, we obtain:

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

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

And so we find:

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