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Find the values of $a$ so that the curves $$y = x^2 + x + a$$ and $$y = x^2 + 1$$ share a tangent line.

Please help, I have been trying to solve this question for the past 1 week but haven't been able to find an ans.

Matti P.
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Amanda
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    So what did you try? What was the approach that you used, and what was the result? – Matti P. Sep 16 '19 at 12:31
  • I assume that the tangent for the first eqn is at point b, the second tangent for second eqn is at point c. I form the two tangent equations and then equate to each other. But I have three unknowns and can't seem to simplify to get a = constant. – Amanda Sep 16 '19 at 12:53

4 Answers4

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As it turns out, there is no restriction on $a$ in order for the two curves to have a common tangent line.

In fact, the equation of the common tangent line is,

$$y= \frac{5-4a}{2}x-a^2+\frac 52 a- \frac{9}{16}$$

So, the answer is:

$a$ can be any real number for the two functions to share a common tangent line.

Edit:

Assume the common tangent line is

$$y=kx+m$$

Plug it into the two curves given to get,

$$x^2+(1-k)x+a-m=0$$ $$x^2-kx+1-m=0$$

Since the line is tangential to the curves, there is only one intersecting point with each curve. So, their respective discriminants are zeros,

$$(1-k)^2-4(a-m)=0$$ $$k^2-4(1-m)=0$$

Then, $k$ and $m$ can be solved.

Quanto
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  • In addition, since they're parabolas, they have the line at infinity as another common tangent. On a side note, how did you find it? I used the dual curves, but if you have a better way I would be interested to see it. Anyway +1. – Jan-Magnus Økland Sep 16 '19 at 15:03
  • Thanks. I solved it by letting zero the discriminants of joint equations. – Quanto Sep 16 '19 at 15:43
  • May I know how you got the equation of the common tangent line? Perhaps you can elaborate a little more on the method – Amanda Sep 17 '19 at 15:10
  • @Amanda - Sure. I added the derivation in the answer. – Quanto Sep 17 '19 at 18:31
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For any $x_0$, the slope of the tangent line to $g(x)= x^2+ 1$ at $(x_0, x_0^2+ 1)$ is $2x_0$ so the tangent line is $y= 2x_0(x- x_0)+ x_0^2+ 1$. That will also be tangent to $f(x)= x^2+ x+ a$ at $(x_1, x_1^2+ x_1+ a)$ if its slope there is $2x_1+ 1= 2x_0$. We have the two equations $2x_0(x_1- x_0)+ x_0^2+ 1= x_1^2+ x_1+ a$ and $2x_1+ 1= 2x_0$.

user247327
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  • Yes, I got this two equations too, but can't simplify to get a= constant. I can only express a in terms of either x0 or x1 – Amanda Sep 16 '19 at 13:10
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It is $$b^2+b+a=m_1b+n$$ $$c^2+1=m_2c+n$$ where $$y=mx+n$$ the tangent line. $$m_1=f'(b),m_2=f'(c)$$

Dr. Sonnhard Graubner
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  • Does this mean I can only find the values of a expressed in either b/c, but cannot find an ans where a is a number? When the question asked for "values of a", I assumed that they wanted actual numbers as the answer. – Amanda Sep 16 '19 at 13:09
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The two parabolas are each other's translates. The vertex of the second is at $(0,1)$, while the first one is at $(-1/2,a-1/4)$. This translation maps only lines with slope $$m = {a-1/4-0 \over -1/2-1} = \frac52-2a$$ onto themselves and the slope of the tangent uniquely determines the point of tangency to a parabola, so if there is a common tangent to the two parabolas, it must also have this slope.

The points at which the tangents to the parabolas have the required slope are, respectively, $p_1=\left(\frac34-a,a^2-\frac32a+\frac{21}{16}\right)$ and $p_2=\left(\frac54-a,a^2-\frac{10}4a+\frac{41}{16}\right)$. We therefore have $${y_2-y_1\over x_2-x_1} = \frac52-2a = m,$$ so the two tangents coincide. This clearly holds for all real $a$.

amd
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