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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.

Steblo
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2 Answers2

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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 $8$.

enter image description here

Math Lover
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  • Thank you. I had got as far as your first line but had failed to find a way to factorise /simplify it as you did. In fact, I follow all your steps but the basic one of getting from line one to line two. I need to concentrate on simplification/factorising. – Steblo Aug 04 '21 at 19:16
  • @Steblo in a lot of these problems such as finding locus, some algebra, rearrangement of equations and factorizations become important. you may want to look into some algebra related exercise material. – Math Lover Aug 04 '21 at 19:22
  • The idea was to convert $(1)$ into a form that had only $pq, p + q$ and $p^2 + q^2$ terms. I used the fact that $(p-q)^2 = p^2 + q^2 - 2 p q = p^2 + q^2 + 2 p q - 4 pq$ $ = (p+q)^2 - 4 pq$. Same with $(p^2 - q^2)^2$ – Math Lover Aug 04 '21 at 19:25
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    Actually, I'm just in the process of working this out for myself! Previously, I've only been accustomed to $(p^2-q^2)=(p-q)(p+q)$. Now I'm applying same technique to ($ap^2-aq^2)$ – Steblo Aug 04 '21 at 19:29
  • That is good and in fact if you do it correctly, it is much simpler than what I did. – Math Lover Aug 04 '21 at 19:35
  • I would never have seen the connections in your comment above:$(p-q)^2=...$ etc. Even if I had sat for hours. I suppose it's one of those connections/tricks you just have to learn. – Steblo Aug 04 '21 at 19:38
  • Let us continue this discussion in chat. – Math Lover Aug 04 '21 at 21:19
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You shoud use formula $d^2=(ap^2-aq^2)^2+(2ap-2aq)^2$.

Put $x=apq$, $y=a(p+q)$ to the left part and formula of $d^2$ to the right part and expand all the parentheses.

Ivan Kaznacheyeu
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