-2

A common tangent to two curves is a line that is tangent to the two curves, but not necessarily at the same point.

Find, in terms of $a$ and $b$, the explicit equation of the common tangent to the two curves $y = x^2 + ax + b$ and $y = x^2 + bx + a$, where $a$ is not equal to $b$.

Also find, in terms of $a$ and $b$, the $x$ coordinates of the points of contact of this common tangent with each of the two curves.

G Cab
  • 35,272
Arrjun Ram
  • 9
  • 1
    did you make any try ? what do you know on this subject ? – G Cab Apr 20 '18 at 18:38
  • 1
    Welcome to math.stackexchange. When stating an exercise, you are expected to explain what you have tried when working the problem and what difficulty you have with the problem. – John Wayland Bales Apr 20 '18 at 18:41
  • I have tried substituting in extra parameters to try and get the common tangent but it ends p being too complex with too many unknowns – Arrjun Ram Apr 20 '18 at 19:20
  • That description of your efforts doesn’t really tell us much about them. There are many ways to attack this problem, and you’re much more likely to get an answer that you’ll understand (or any answer at all, for that matter) if you show your attempt explicitly in your question. – amd Apr 20 '18 at 19:29
  • What I have done so far is differentiated both of the equations to get the gradient function (2x+a) and (2x+b). Then substituted the points of x=c into the first equation to get coordinates of (c,c^2+ac+b) and (d,d^2+bd+a). From there worked out the gradient of the tangent by doing the change in y over the change in x. That's as far as I have reached. – Arrjun Ram Apr 20 '18 at 19:41
  • the tangent line would touch the two curves at two points (one point on each curve) and the slope of each curve at the respective point would be the same as the slope of the tangent line. I wrote it in a convoluted way, instead of trying to fix it now, it might be better if you try to figure what I meant and how to use it :) Do you know that the slope is the same as the derivative? Could you find the derivative of $y = x^2 + ax + b $ ? – Mirko Apr 20 '18 at 19:41
  • Yes I get what you mean, thank you, but I'm still unsure what to do after the stage of getting the points and the gradients as so far I have two points (c,c^2+ac+b) and (d,d^2+bd+a) and doing the change in y over change in x, I get (d^2 + bd + a - c^2 - ac- b)/d-c – Arrjun Ram Apr 20 '18 at 19:47
  • A line through points $(x_1,y_1)$ and $(x_2,y_2)$ has slope $m=\frac{y_2-y_1}{x_2-x_1}$ (as you already indicated, change in y over change in x) and equation $y-y_1=m(x-x_1)$. – Mirko Apr 20 '18 at 19:50
  • Okay that makes sense to get the equation. But how would I find the points y1 and x1, do I use algebra or substitute in numbers? – Arrjun Ram Apr 20 '18 at 19:52

3 Answers3

1

Calling

$$ \left\{ \begin{array}{rcl} f_1(x,y) & = & x^2 + a x + b - y\\ f_2(x,y) & = & x^2 + b x + a - y\\ L& \rightarrow & p = p_1+ \lambda (p_2-p_1) \end{array} \right. $$

with $p = (x,y), \ p_1 = (x_1,y_1) \ p_2 = (x_2,y_2)$ such that $f_1(p_1) = f_2(p_2) = 0$ the tangency problem can be established as:

$$ \left\{ \begin{array}{l} f_1(p_1+\lambda(p_2-p_1)) = 0 \Rightarrow ((a + 2 x_1) (x_1 - x_2) - y_1 + y_2)^2 -4 (x_1 - x_2)^2 (b + x_1 (a + x_1) - y_1)= 0\\ f_2(p_1+\lambda(p_2-p_1)) = 0\Rightarrow ((b + 2 x_1) (x_1 - x_2) - y_1 + y_2)^2 -4 (x_1 - x_2)^2 (a + x_1 (b + x_1) - y_1) = 0 \end{array} \right. $$

because in the corresponding equations due to $f_i = \phi(x_i,y_i)\lambda^2+\psi_i(x_i,y_i)\lambda + \theta_i(x_i,y_i) = 0, \ \ \{i = 1,2\}$ the discriminant associated to the $\lambda$ solution should be null.

Now substituting $y_1 = x_1^2 + a x_1 + b$ and $ y_2 = x_2^2+b x_2 + a$ we obtain

$$ \left\{ \begin{array}{rcl} b - (x_1 - x_2)^2 + a (x_2-1) - b x_2 & = & 0\\ b + a (x_1-1) - b x_1 + (x_1 - x_2)^2 & = & 0 \end{array} \right. $$

and solving for $x_1,x_2$ we get

$$ \left\{ \begin{array}{rcl} x_1 & = & 1/4 (4 - a + b)\\ x_2 & = & 1/4 (4 + a - b) \end{array} \right. $$

and the tangent line reads

$$ L\rightarrow p = \left\{\frac{1}{4} (-a+b+4),\frac{1}{16} \left(-3 a^2+2 a (b+4)+b (b+24)+16\right)\right\}+\lambda \left\{\frac{a-b}{2},\frac{1}{4} (a-b) (a+b+4)\right\} $$ Attached a plot for $a = 2, b = -2$ enter image description here

Cesareo
  • 33,252
0

$$y_1=x_1^2+ax_1+b \tag{1}$$

$$y_2=x_2^2+bx_2+a \tag{2}$$

Equating the slope of the common tangent:

$$m=2x_1+a=2x_2+b$$

$$x_1=\frac{m-a}{2} \tag{3}$$

$$x_2=\frac{m-b}{2} \tag{4}$$

Now \begin{align} m &= \frac{y_1-y_2}{x_1-x_2} \\ &= \frac{x_1^2+ax_1+b-x_2^2-bx_2-a}{x_1-x_2} \\ &= \frac{(m-a)^2+2a(m-a)+4b-(m-b)^2-2b(m-b)-4a}{2(m-a)-2(m-b)} \\ &= \frac{m^2-a^2+4b+b^2-m^2-4a}{2(b-a)} \\ &= \frac{b^2-a^2+4(b-a)}{2(b-a)} \\ &= \frac{a+b}{2}+2 \end{align}

Ng Chung Tak
  • 18,990
  • Thank you, once you've worked out the gradient as you have how would you then get the explicit equation of the common tangent to the two curves and the coordinates of the points of contact of this common tangent with each of the two curves. – Arrjun Ram Apr 20 '18 at 20:10
0

Before grinding through some generic calculation, it’s a good idea to look for special features of the problem that will simplify it. In this case, the two curves are parabolas that are related by a translation. Any common tangent line must be invariant under this translation, which implies that it must be parallel to the translation direction.

The vertices of the two parabolas are easily found to be at $\left(-\frac a2,b-\frac{a^2}4\right)$ and $\left(-\frac b2,a-\frac{b^2}4\right)$. The difference is $\left(\frac12(b-a),\frac14(b-a)(a+b+4)\right)$ from which we get $\frac12(a+b+4)$ for the slope of the common tangent. From here it’s a fairly straightforward matter of finding where on the curves the derivative has this value and generating the equation of the line from a point and slope or from the two points.

amd
  • 53,693