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Here is the problem: Determine $c$ so that the straight line joining $(0, 3)$ and $(5, -2)$ is tangent to the curve $y = c/(x + 1)$.

So, what I know: the line connecting the points is $y=-x+3$. I need to find a value for c such that the derivative of $c/(x+1)$ can be $-x+3$. I took the derivative of $c/(x+1)$, and found a line via point-slope form, so $$y-3=-\frac{cx}{(x+1)^2} \quad \Rightarrow \quad y=-\frac{cx}{(x+1)^2} + 3 $$

At this point the tangent line is $y=-x+3$. So, $-x = (-cx/(x+1)^2)$. From there, I can conclude $c=x^2+2x+1$. How do I conclude, without looking at a graph, what value of $x$ to choose? I know from the answer that at $x=1$, $c=4$, and the tangent line meets the curve at that point $x=1$. But how would I determine this?

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from $\frac{c}{x+1} = -x+3$ and $-\frac{c}{(x+1)^2} = -1$ , you got $x+1=-x+3$

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Hint:

Here are two curves whose slope are supposed to be equal: $$ y=-x + 3 $$ and $$ \frac{c}{x+1}=y $$

equating the slopes of both of curves,

we have $$ -1=\frac{-c}{(x+1)^2} $$

this gives you $$ (x+1)^2=c\implies x=-1\pm \sqrt c $$

note that from here we need $c$ should be non-negative. i.e. $c\geq 0$

Now at $x=-1\pm \sqrt c$ both curves touch each other. So, $$ -x + 3=\frac{c}{(x+1)} $$ at $x=-1\pm \sqrt c$.

Solve this equation replacing $x=-1\pm \sqrt c$. and you will have value of $c$ you want.

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  • $\begingroup$ thx. The key step, I think, is equating the slope of both curves. $\endgroup$ – Chas Feb 16 '15 at 7:35
  • $\begingroup$ Always Welcome :) $\endgroup$ – Harish Feb 16 '15 at 23:22
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A line will either intersect the curve at more than one points(intersection), or it will touch at a single point(tangent), or it wont intersect at all. In all the cases if you solve the equations of the line and the curve simultaneously you will get the solutions which reflect the above three conditions.

Now you straight line is $$y=-x+3$$and the curve is $$y=\frac{c}{x+1}$$ which is a hyperbola. Solve these two equations to get the intersection point(s).

$$-x+3=\frac{c}{x+1}$$ $$x^2 -2x+(c-3)=0$$ Solve the equation to get $$x=1\pm\sqrt{4-c}$$ therefore $$y=2\mp\sqrt{4-c}$$ The points of intersection of these two curves are $A(1+\sqrt{4-c},2-\sqrt{4-c})$ and $B(1-\sqrt{4-c},2+\sqrt{4-c})$. Now the thing is a line and a hyperbola can intersect at maximum 2 points and those 2 points are known. Since the line is a tangent to the hyperbola, we can equate both the coordinates. In other words, the line should touch the hyper bola at one point for that both the points must have the same value. Equating the x and y coordinates of A and B respectively we will get: $$1+\sqrt{4-c}=1-\sqrt{4-c}$$ $$1+\sqrt{4-c}=1-\sqrt{4-c}$$ Only one condition: $$\sqrt{4-c}=0$$ which means $c=4$ and hence, if you put this back in the value of $x$ you get only one point $x=1\pm\sqrt{4-c}=1\pm\sqrt{0}=1$ which proves that actually the line is a tangent.

The $y$ value is $y=2$ So the point of contact is $(1,2)$

Now another check whether our calculation is correct or not, calculate the slopes separately of the line and hyperbola at the contact point $(1,2)$. Slope of line at $(1,2)$ is $-1$. Slope of line passing through the point $(1,2)$ is :

$$\frac{dy}{dx}|_{(x=1,y=2)} = -\frac{c}{(x+1)^2}=-\frac{4}{4} = -1$$

So we verified that the line which touches the hyperbola has the same slope as that of the line that we came up with.

Check this online plotting tool https://www.geogebra.org/3d?lang=en put the input and output functions to see the plot it will help you visualize it.

Hope this helps ...

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