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Tangents drawn from the point $(\alpha,\alpha^2)$ to the curve $x^2+3y^2=9$ include an acute angle between them, then find $\alpha$.

My attempt is by using the equation for pair of tangents from an external point to the curve. If the curve is $S=0$ then the equation for pair of tangents is $SS_1=T^2$, where $S_1=0$ is obtained after putting the external point in $S=0$. And $T=0$ is obtained after changing $x^2$ to $xx_1$ and $y^2$ to $yy_1$ in $S=0$, where $(x_1,y_1)$ is the external point from which the tangents are being drawn.

So,

$$SS_1=T^2$$ $$\implies (x^2+3y^2-9)(\alpha^2+3\alpha^4-9)=(x\alpha+3y\alpha^2-9)^2$$

Now, if $\theta$ is the angle between the tangents then $\tan\theta=|\frac{2\sqrt{h^2-ab}}{a+b}|$, where, $2h=$ coefficient of $xy=6\alpha^3, a=$ coefficient of $x^2=3\alpha^4-9$ and $b=$ coefficient of $y^2=3\alpha^2-27$

Therefore, $\tan\theta=|\frac{2\sqrt{9\alpha^6-9(\alpha^4-3)(\alpha^2-9)}}{3\alpha^4+3\alpha^2-36}|=|\frac{2\sqrt{\alpha^6-(\alpha^6-9\alpha^4-3\alpha^2+27)}}{\alpha^4+\alpha^2-12}|=|\frac{2\sqrt{9\alpha^4+3\alpha^2-27}}{\alpha^4+\alpha^2-12}|$

Not able to proceed next.

aarbee
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  • What is $\alpha$ in the cited problem? What does it mean that "tangents include an acute angle? (And if they don't, do we still have to find that $\alpha$?) What are $S,S_1,T$ in the attempt? – dan_fulea May 04 '21 at 17:44
  • @dan_fulea I have made the edit. Also, from the external point, two tangents would be drawn to the ellipse. And those tangents would have an acute angle between them. – aarbee May 04 '21 at 17:51
  • Do you mean that, if $A=(\alpha,\alpha^2)$ if $T$ and $T'$ are the such that the lines $AT$ and $T'$ are tangent to the ellipse, then the angle $\angle T\hat AT'$ is acute? – José Carlos Santos May 04 '21 at 17:53
  • @JoséCarlosSantos Yes. – aarbee May 04 '21 at 17:54
  • Where did this equation $SS_1=T^2$ come from? Please explain. – Ted Shifrin May 04 '21 at 17:59
  • Please take time, and edit the question so that objects are introduced in the right order. Please say what is that $\alpha$ before using it. The include (associated to an angle) is still unclear. Because the inclusion makes sense for sets, and some "angle" seen as sector maybe can "include" some other angle". Then start also defining that $S$ and $S_1$ immediately, not in some later EDIT. That $T^2$ still makes no sense. The implication makes no sense, since $S,S_1,T$ have no relation to $x,y,\alpha$ as they stay there. – dan_fulea May 04 '21 at 17:59
  • @dan_fulea ok, 2 mins. – aarbee May 04 '21 at 18:00
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    From experimentation in geogebra there's a range $\alpha> 1.7$ (and $\alpha< -1.7$) for which the angle is acute. – Jan-Magnus Økland May 04 '21 at 18:03
  • @dan_fulea I have explained my solution in order now. Also, the external point is $(\alpha,\alpha^2)$. I have not defined it. It is part of the question. Also, if there are two intersecting lines then they would have an obtuse angle between them and an acute angle. The word 'include' might be wrongly used here, don't know. It's part of the question. – aarbee May 04 '21 at 18:09
  • @TedShifrin that's the equation for pair of tangents from an external point to a curve. I have edited my post. Hopefully things are clearer now. – aarbee May 04 '21 at 18:17
  • As it stays no, the problem is "Tangents drawn from the point $(\alpha,\alpha^2)$..." and here we stop. What is $\alpha$? It seems that the problem is copied / translated from somewhere, as it stays it makes no sense from here, where i am reading it. "Find $\alpha$" is also unclear. But ok. I will let it as it is without downvote and quit here. – dan_fulea May 04 '21 at 18:20
  • @dan_fulea The external point could be $(2,4)$ or $(5,25)$ or $(\sin4, \sin^24)$. I don't know. Maybe the question setter should have written that $\alpha$ is a real number. But I guess even without that, things are clear enough to start the problem. – aarbee May 04 '21 at 18:26
  • @aarbee pay attention to the comment by Jan-Magnus. An infinite number of values $\alpha$ satisfy. – Viera Čerňanová May 04 '21 at 18:46
  • @user376343, thanks for pointing this out. – aarbee May 04 '21 at 19:15
  • @Jan-Magnus, thanks. – aarbee May 04 '21 at 19:15
  • $\theta<90°$ for $\alpha>\sqrt3$. – Intelligenti pauca May 05 '21 at 13:41
  • @Intelligentipauca Thanks. Would you mind elaborating this in an answer so that I could understand the underlining reasoning better? Thanks. – aarbee May 05 '21 at 20:43

2 Answers2

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Hint: $\alpha^4+\alpha^2-12=(\alpha^2-3)(\alpha^2+4)$ making the angle $\pi/2$ for $\alpha=\pm\sqrt{3}.$

Now check for say $\alpha\approx 2.2$ parabolaellipsetangentsangle

Jan-Magnus Økland
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  • Thanks for the wonderful hint but I am still not able to conclude things. In the comments above you have written that for $\alpha\gt\sqrt3$ and $\alpha\lt-\sqrt3$, angle would be acute. How did we decide that? I understand the denominator would be positive for the above range and would be negative otherwise. Does that have anything to do with acuteness or obtuseness of the angle? How? Any other hint would be welcomed, thanks. – aarbee May 05 '21 at 20:25
  • @aarbee You need to check that the square root exists (which it does outside the ellipse), and then it is clear that as $\alpha$ varies the angle changes from obtuse to acute at $\sqrt3$. Then it’s a matter of checking in which direction. Outside the ellipse but close to it the angle is obtuse as you can see by drawing it, or thinking that at the ellipse you start with a double tangent which has angle $\pi$. Far from the ellipse the angle is acute as you can check for a value or by using that the ellipse is bounded and the situation looks like chopsticks about an egg. – Jan-Magnus Økland May 06 '21 at 01:26
  • When I put $\alpha=0$, square root become negative, so, not possible. When I put $\alpha=1$ or $-1$, square root exists, but $(1,1)$ and $(-1,1)$ sure lie inside the ellipse. So, can we relate the existence of square root with the point being outside the ellipse? How do we find points which lie outside the ellipse? Also, I understand if the point is just outside the ellipse then the angle will be obtuse. But how do we put a numerical figure to it? I understand at $\alpha=\pm\sqrt3$, angle will be 90, but not able to decide how to go left and right of it. (Thankyou for your engagement.) – aarbee May 06 '21 at 06:47
  • @aarbee $h^2-ab=27(3\alpha^4+\alpha^2-9)$ which has zeros for $\alpha\approx\pm 1.25435$ which does fit with the border of the ellipse. – Jan-Magnus Økland May 06 '21 at 07:01
  • Thanks, I had made a mistake in calculating $h^2-ab$, I have edited my post now. But I am sorry I am still not able to see why there would an acute angle for $\alpha\gt\sqrt3$. – aarbee May 06 '21 at 07:11
  • @aarbee See the edit. – Jan-Magnus Økland May 06 '21 at 07:33
  • Got it now, thankyou so much. – aarbee May 06 '21 at 07:48
1

$$y=mx+c$$ will be tangent of $$\dfrac{x^2}{3^2}+\dfrac{y^2}3=1$$ if $c^2=9m^2+3$

$$(y-mx)^2=c^2=9m^2+3$$

$$ m^2(9-x^2)+2mxy+3-y^2=0$$

As these pass through $(\alpha,\alpha^2),$

$$ m^2(9-\alpha^2)+2m\alpha^3+3-\alpha^4=0$$ which is a quadratic equation in $m$

$$m_1+m_2=-\dfrac{2\alpha^3}{9-\alpha^2}, m_1m_2=\dfrac{3-\alpha^4}{9-\alpha^2}$$

If the angle between the tangents be $p,$

$$\tan p=\left|\dfrac{m_1+m_2}{1-m_1m_2}\right|$$

lab bhattacharjee
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  • But that's not giving us an answer. Or is it? – aarbee May 04 '21 at 17:55
  • The problem is still unclear, but ok, we now have a solution, which is immediately edited. I am just curious: Did the above find that $\alpha$? Where is it? – dan_fulea May 04 '21 at 17:55
  • @dan_fulea yes, the above solution also gives $\tan\theta$ in terms of $\alpha$, the way I got. But we haven't got $\alpha$ yet. – aarbee May 04 '21 at 17:59
  • So just look at the question as it is now. Is that a question? Is that a kind of something having any sense? Now look at your answer. Is there any starting point? Do we have some sentence with a clear meaning? Which is the subject in "As these pass through $(\alpha,\alpha^2)$..." And where do we know that "that stuff" is passing through? You immediately invent $m_1,m_2$, and relate them to some tangents we never saw before, but ok, their angle is $p$, and now we can write some equality connecting the $p$, never known to my eyes, to something that can be written only in terms of $\alpha$... – dan_fulea May 04 '21 at 18:04
  • @dan_fulea, lab bhattacharjee's solution is a fairly acceptable solution. Everything he has written is in standard format. – aarbee May 04 '21 at 18:12