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