I am struggling with the last part of this question.
Prove that the tangent at the point $(at^2,2at)$ on the parabola $y^2=4ax$ has the equation $ty=x+at^2$. Find, in their simplest form, the coordinates of T, the point of intersection of the tangents at the points $P(ap^2,2ap)$ and $Q(aq^2,2aq)$ on the parabola $y^2=4ax$. If PQ is of constant length $d$, show that T lies on the curve whose equation is:
$(y^2-4ax)(y^2+4a^2)=a^2d^2$
My workings for the first parts are as follows:
$x=at^2; y=2at$
$\frac{dy}{dx}=\frac{dy}{dt}/\frac{dx}{dt}=\frac{1}{t}$
$\rightarrow y-2at=\frac{1}{t}(x-at^2)\rightarrow ty=x+at^2$ (1)
Equation of tangent at P: $py=x+ap^2$
Equation of tangent at Q: $qy=x+aq^2$
At T: $\frac{x+ap^2}{p} = \frac{x+aq^2}{q} \rightarrow qx = ap^2q= px+aq^2p \rightarrow x=apq$
from (1) $y=\frac{x+ap^2}{p} \rightarrow y=a(p+q)$
So coordinates of T are $apq, a(p+q$
Then comes this part which I cannot solve:If PQ is of constant length $d$, show that T lies on the curve whose equation is:
$(y^2-4ax)(y^2+4a^2)=a^2d^2$
I am assuming the calculations I have already done will help me but I cannot see the link.
Best Answer
I simplified the answer further.
$PQ^2 = d^2 = (ap^2 - aq^2)^2 + (2ap - 2aq)^2$
$d^2 = a^2 (p+ q)^2 (p-q)^2 + 4 a^2 (p - q)^2$
$d^2 = a^2 (p-q)^2 ((p+q)^2 + 4)$
Using $(p-q)^2 = (p+q)^2 - 4 pq \ $,
$a^2d^2 = (a^2(p+q)^2-4a^2pq) (a^2(p+q)^2 + 4a^2) \tag1$
Now as you found, the coordinates of $Q$ is,
$x = apq \implies pq = \cfrac{x}{a}, y = a (p+q) \implies p + q = \cfrac{y}{a}$
Plugging in values of $pq$ and $p + q$ into $(1)$, you get the locus in desired form immediately.
Taking $d = 8, a = 1$, here is a diagram that shows the locus. Length of the chord of contact from any point on this curve to the parabola will be $8$.