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We have the parabolas: $y = 2x^{2} + 2x + 1$ and $y = 2x^{2} - 2x - 1$.

Finding the common tangents with calculus is as straight forward as just solving a system of equations with the derivatives of both parabolas and $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$; but how about in analytical geometry?

A.Γ.
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Johnny Bueti
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  • how do you find tangents to curves in analytical geometry? – Sonal_sqrt Jun 08 '18 at 06:37
  • It's pretty easy if you're given a point; you would find the interception point of the curve and and arbitrary line that passes through the given point. You could also just imply the delta of the parabola is equal or greater than zero and find out for which value that is true. In this case, however, you would be left out with an equation with two parameters. – Johnny Bueti Jun 08 '18 at 06:41
  • You will have to use calculus because slope of tangent is always going to be a limit. You can avoid using derivative formulas if you calculate slopes using limits definition of derivatives. But I think that is about as far as we can push it. – Sonal_sqrt Jun 08 '18 at 06:46
  • Which is definitely odd since this kind of problem has been given to be solved at third grade students. Not to sound arrogant, and maybe a bit off topic, but as I thought, this is completely crazy. A third grader (high school) can't solve this. – Johnny Bueti Jun 08 '18 at 06:49
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    Is third grade high school the last year in school before college? I will really depend what all the person trying to solve this knows. If this was taught in curriculum then he should be able to solve this. In India this is something that precollege students are able to solve. – Sonal_sqrt Jun 08 '18 at 06:52
  • Nope, it's the second last year before college. Limits, derivatives, integrals and calculus in general hasn't been taught to them yet, not even trigonometry; it's still plain analytical geometry. – Johnny Bueti Jun 08 '18 at 06:55
  • Then it might not be possible to solve this. And you are saying conic sections are taught before trigonometry and limits. This is very strange. – Sonal_sqrt Jun 08 '18 at 07:00
  • Italy for you my man. Top-notch mathematics education. Conic sections (parabola, hyperbola, ellipse and circumference) are taught before limits and trigonometry, and after sets and functions. – Johnny Bueti Jun 08 '18 at 07:07
  • But students can do some kind of reverse engineering, can't they? If you have parabola $y=f(x)$ and want to find a tangent at $(a,b)$, then all you need to do is solve for $m$ for which $y-b=m(x-a)$ meets $y=f(x)$ at exactly one point $(x,y)=(a,b)$. The knowledge of the discriminants of quadratic equations should be very handy in this situation. This is the only way I know in order to avoid calculus. – Batominovski Jun 08 '18 at 07:15
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    Hint: The parabolas are translations of each other, so any common tangents are invariant lines of the translation. That means there’s only one common tangent line, and you can compute its slope. – amd Jun 08 '18 at 07:29
  • Exactly, but you are given no information whatsoever about the tangent. Even forcing both discriminants to be greater or equal than zero after intersecting the parabolas and the general equation of a line, you are left out with infinite tangents. – Johnny Bueti Jun 08 '18 at 07:37
  • Take two lines $ y= mx+c, y= Mx+C$ cutting each parabola separaately. Make discrimnants vanish for each line solution and then equate $x$ term and constant.to merge them.as a single straight line. – Narasimham Jun 08 '18 at 08:49

5 Answers5

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A line $y=kx+m$ is a tangent to the parabola $y=ax^2+bx+c$ iff the quadratic equation $$ kx+m=ax^2+bx+c $$ has exactly one solution, which makes the discriminant to be zero. In our case $$ kx+m=2x^2+2x+1\iff 2x^2+(2-k)x+1-m=0,\\ kx+m=2x^2-2x-1\iff 2x^2-(2+k)x-1-m=0 $$ should each have a unique solution $x$, that is the discriminants are zeros for both equations $$ \begin{cases} (2-k)^2-8(1-m)&=&0,\\ (2+k)^2+8(1+m)&=&0. \end{cases} $$ Solve the system: $k=-2$, $m=-1$.

A.Γ.
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This pair of parabolas are meant to make the problem easy. Complete the square to get $$2y = 4x^2 + 4x + 2 = (2x+1)^2 + 1 = 4(x + \frac12)^2 - 1 + 2$$ for the first parabola and $$2y = 4x^2 - 4x -2 = (2x-1)^2 - 3 = 4(x - \frac12)^2 - 1 - 2$$ for the second. The vertex of the first is at $\,(-\frac12,\frac12)\,$ and the second at $\,(\frac12,-\frac32).\,$

The line between them has slope $\,-2,\,$ which is the slope of the common tangent line. The parabolas intersect at the difference $\, 4x + 2 = 0 \,$ which is $\, x = -\frac12. \,$ Now the line with slope $\, -2 \,$ that passes through the intersection is $\, y = -2x + \frac12.\,$ That line intersects the second parabola at solutions to $\, 2x^2 - \frac32 = 0 \,$ but the average of the two solutions is $\, x = 0. \,$ The point on the second parabola with this $\,x\,$ value is $\, (0,-1) \,$ and is the point of tangency of the parabola with the line $\, y = -2x - 1. \,$

To check, we find that the line $\, y = -2x + \frac12\,$ intersects the first parabola at solutions to $\, 2x^2 + 4x + \frac12 = 0. \,$ and the average of the two solutions is $\, x = -1.\,$ The point on the first parabola with this $\,x\,$ value is $\, (-1,1) \,$ which is the point of tangency with the line $\, y = -2x - 1. \,$

To summarize, think of the special case of $\, y = (x-1)^2 \,$ and $\, y = (x+1)^2. \,$ The line $\, y = 1 \,$ passes through their intersection point $\,(0,1)\,$ and also another point on each parabola. The line $\, y = 0 \,$ is the common tangent. The geometry of this special case is affinely equivalent to the general case of any two parabolas which are translates of each other.

Somos
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1

Refer to this question here for the equation of a tangent to a parabola.

First parabola: $$y-\tfrac 12=2(x-\tfrac 12)^2$$ Tangent at $(h, k= 2h^2+2h+1)$ : $$\tfrac 12 ((y-\tfrac 12)+(k-\tfrac 12))=2(x+\tfrac12)(h+\tfrac12)\\ y=2(1+2h)x+1-2h^2\qquad (1)$$

Second parabola: $$y+\tfrac 32=2(x-\tfrac 12)^2$$ Tangent at $(m, n=2m^2-2m-1)$: $$\tfrac 12 ((y+\tfrac 32)+(m+\tfrac 32))=2(x-\tfrac12)(m-\tfrac12)\\y=2(2m-1)x-1-2m^2\qquad (2)$$ Equating coefficients in $(1), (2)$ gives $(m,n)=(0,-1)$ and $(h,k)=(-1,1)$.

Hence, from either $(1)$ or $(2)$, equation of common tangent is $$\color{red}{y=-2x-1}$$

Hypergeometricx
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The common tangent is parallel to the line joining the two vertices, hence its equation is of the form $y=-2x+k$.

To find $k$ we can use the fact that this tangent has only one point in common with any of the parabolas (the second one, for instance). That is, the system $$ \cases{y=-2x+k\\ y=2x^2-2x-1 } $$ must have only one solution.

Eliminating $y$ we find $2x^2-1=k$, which has a single solution only if $k=-1$.

Intelligenti pauca
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The equation of tangent to y=$ax^2+bx+c$ is \begin{equation} y-mx=c-\frac{{(m-b)}^2}{4a} \end{equation}

The common tangent to y= $2x^2+2x+1$, y= $2x^2-2x-1$ must have the same slope and y-intercept. Therefore,

\begin{equation} c_1-\frac{{(m-b_1)}^2}{4a_1}=c_2-\frac{{(m-b_2)}^2}{4a_2} \end{equation}

Here, $c_1$= 1, $c_2$= -1, $a_1$= $a_2$= 2, and $b_1$= 2, $b_2$= -2. Using these values we can get the value of m that satisfies the last equation and is equal to -2. Thus, the equation of tangent becomes, y+2x= -1.

L Manimaran
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