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Full Question: A variable parabola of fixed latus rectum 4b and having axis parallel to x–axis, lies completely in Ist and IVth quadrant and cuts the fixed parabola $y^2=4ax$ orthogonally. The locus of vertex of the variable parabola is (and on what inputs is it valid)?

My attempt:

The variable parabola was assumed as $(y-k)^2=4b(x-h)$ , the primary being $y^2=4ax$. Let $(at^2,2at)$ be point of intersection. I found the derivatives at the desired point.

let $f'(x)=1/t$ and $g'(x)=\frac{2b}{2at-k}$

Putting the product as negative 1, we get $2at^2-tk+2b=0$

I say that only one t exists, then $D=0$ which gives me the final result $|y|=4\sqrt{ab}$

BUT when I use it as the locus of the vertex. it only satisfies for a unique value, i.e. $x=0$. My question is how do I prove that it's only possible for $x=0$. My provided solution maybe wrong, feel free to use your method as well.

Anindya Prithvi
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  • Why should there be only one $t$? – user10354138 Jul 24 '20 at 14:29
  • Because if there's more than one. the curve cant exist. Try using desmos and see the point of intersections. I have tried this and tested as well. If you need help/link to graph, I will try to provide it soon. – Anindya Prithvi Jul 24 '20 at 14:37
  • Assuming that there should only be one $t$, doesn't the method by which you got $x=0$ say that it is only possible for $x=0$? I didn't get what you are asking if you have got $x=0$. Could you share how you got $x=0$? – Sameer Baheti Jul 24 '20 at 14:45
  • $D=0$ means the discriminant for the given eqn is zero. So, when you substitute k as y you get mod y= 4 $\sqrt(ab)$ as the locus, The thing is, when you test the locus for its validity, you only get 1 point where it is valid. I will put a desmos link https://www.desmos.com/calculator/urxrik8mz9 – Anindya Prithvi Jul 24 '20 at 15:14
  • @Aretino Please reconsider your comment and my supplied solution. D=0 is essential for the proof unless you do it without parametric form (which i would be very pleased to know because initially i tried that way as well) – Anindya Prithvi Jul 24 '20 at 16:24
  • So do you want to see how to get $x=0$ mathematically from where you have reached? – Sameer Baheti Jul 24 '20 at 16:26
  • @SameerBaheti yes,that's the last line of the body of the question as well. I want a proof that y=4$\sqrt(ab)$ is true only when vertex is at x=0 – Anindya Prithvi Jul 24 '20 at 16:28
  • Content defacing (unwarranted/revenge downvote) attribute to user:730361 – Anindya Prithvi Dec 06 '20 at 19:24

1 Answers1

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Imposing only one solution for $t$ is not requested: it may well happen that for a given value of $k$ two different values of $t$ (and thus of $h$) be found.

Your approach is correct, but you forgot that your goal is that of finding vertex $(h,k)$ as a function of $t$, and from your equation $2at^2-tk+2b=0$ one immediately gets: $$ k=2at+{2b\over t}. $$ To find $h$ just use $$ h=x-{(y-k)^2\over4b}=at^2-{b\over t^2}. $$ These are then the parametric equations of the locus.

Notice that the parabolas, if $a\ne b$, also have another intersection point, where they are not orthogonal: probably the text of the problem should have mentioned that, to avoid confusion.

EDIT.

You can see below a diagram, with $a=1$ (black parabola) and $b=4$ (red parabola). Parabolas meet at $P=(4,4)$, where they are orthogonal (dashed lines are the tangents at $P$), but they also meet at $Q$. Green curve is the locus of vertex $V$ as $P$ varies on the black parabola: as you can see, for a given value of $k$ there are in general two possible values for $h$.

Note also that focus $F'$ of the variable parabola lies on line $PF$ (this is required by orthogonality). It is in fact possible to construct focus $F'$ (and vertex $V$) in a purely geometrical way.

enter image description here

Intelligenti pauca
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  • I dont understand how imposing one solution gets you a tangent. If there are more than one t's the plot for the parabola would extend to 2,3 quadrants as well, and if it's stretched in quadrant 1,2, the parabola becomes a hyperbola. Would be helpful if the tangency is explained. (h,k) seems correct and plots well. Please provide cartesian form. There is no "Another point of intersection". Please verify your claims. Verify here if required https://www.desmos.com/calculator/bbshcdsaqy – Anindya Prithvi Jul 24 '20 at 17:20
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    @AnindyaPrithvi I added a diagram to show you. – Intelligenti pauca Jul 24 '20 at 17:38
  • Please add the condition of tangency you were talking about in comments. The cartesian form may not be posted. – Anindya Prithvi Jul 24 '20 at 18:14
  • Rest of the answer is very much clear, thanks a lot. – Anindya Prithvi Jul 24 '20 at 18:14
  • @AnindyaPrithvi You are right about tangency: requesting $D=0$ has nothing to do with tangency of parabolas. I corrected and added the right explanation. – Intelligenti pauca Jul 24 '20 at 18:18
  • Got itt :) Thank you – Anindya Prithvi Jul 24 '20 at 18:23
  • But still, I don't think we can't request $D=0$ for $2at^2-tk+2b=0$ because, for the same $k$, there are two possible variable parabolas in general. Hence two distinct solutions. Do we agree on this? – Sameer Baheti Jul 24 '20 at 18:36
  • @SameerBaheti No one disagreed on that, its a part of solution, but not the complete set. Check the desmos link in these comments, you'll see the parametric solution covers for my solution as well. – Anindya Prithvi Jul 24 '20 at 19:48
  • @AnindyaPrithvi If you think that $D=0$ is a must, then $h=b-a$ is also a must which is clearly not the case. $${}$$Since they intersect in the point $(at^2,2at)$, where $t=\frac{k}{4a}, k=4\sqrt{ab}\Rightarrow t=\sqrt{\frac ba}$, $$(2at-k)^2=4b(at^2-h)\Rightarrow (2\sqrt{ab}-4\sqrt{ab})^2=4b(b-h)\Rightarrow h=b-a$$ – Sameer Baheti Jul 25 '20 at 04:23
  • How did you get t=k/4a – Anindya Prithvi Jul 25 '20 at 06:07
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    @AnindyaPrithvi By solving $2at^2-2kt+2b=0$ and applying $D=0$, which was correct according to you earlier. I am just pointing out a possible mistake you made earlier. – Sameer Baheti Jul 26 '20 at 09:00