Find the point of tangent given f and the area of a triangel.

[Randyyy]

Hey, the question is as follows:

To the curve [math]y=2-(x-2)^2[/math] both a tangent and a normal is drawn from the same point. The tangent and the normal together shape a triangle with the area of 4.5 area units. Find the point of tangent.

I initially solved it by first expressing a function for both the tangent and the normal as follows:
[math]y_t-f(a)=f'(a)(x-a)[/math][math]y_n-f(a)=-\frac{1}{f'(a)}(x-a)[/math]. Plugging in and simplifying I get that;
[math]y_t(x) =a^2+4x-2ax-2[/math][math]y_n(x) =\frac{8-21a+12a^2-2a^3+x}{2(a-2)}[/math]
If I then find the point at which they intersect the x-axis and take the difference, I have got my base and the height is given by f(a). So a can then be given using area of a triangle.
[math]y_t=0 \implies x_t=\frac{a^2-2}{2(a-2)}[/math][math]y_n=0 \implies x_n=-8+21a-12a^2+2a^3[/math]
The area is given by [math]\frac{bh}{2}=\frac{\mid x_t-x_n \mid f(a)}{2}=\frac{9}{2} \iff a_1=1.73499, a_2=2.26501[/math]a_1 does not yield a solution to the problem, so only a_2 is a valid value. From here it is easy to find the (x,y) coordinates and even the equation for the tangent and normal.

I was then also presented by a solution that used tan(x) and tan(90-x) due to the right angle formed at the top of the triangle (the point where the normal and tangent are drawn from).This piqued my interest, are there other methods/tricks to solving this problem?

 

[Dr.Peterson]

To the curve \(y=2−(x−2)^2\) both a tangent and a normal is drawn from the same point. The tangent and the normal together shape a triangle with the area of 4.5 area units. Find the point of tangent.

The problem isn't stated clearly. Two lines can't form a triangle! It appears that you are taking it this way:

To the curve \(y=2−(x−2)^2\) both a tangent and a normal are drawn from the same point on the curve. The tangent and the normal and the x-axis together form a triangle with the area of 4.5 area units. Find the point of tangency.

Your method looks reasonable. But I don't think it's quite right (assuming I've interpreted the problem correctly). I would expect to have at least two solutions that are symmetrical about the line of symmetry of the parabola; why do you reject your first solution?

I would also not be surprised to see two more solutions with the point below the x-axis, as the area is still positive though the "height" as you are taking it will be negative. I haven't yet tried to solve the entire problem myself.

I was then also presented by a solution that used tan(x) and tan(90-x) due to the right angle formed at the top of the triangle (the point where the normal and tangent are drawn from). This piqued my interest, are there other methods/tricks to solving this problem?

There usually are multiple methods. It would be most useful to you if you made an attempt to find others yourself!

But I have no idea what this second method you saw is, since clearly x is not the x-coordinate, but some angle. Could you show us what was actually done?

 

[jonah2.0]

Beer induced ramblings follow.

...
The area is given by [math]\frac{bh}{2}=\frac{\mid x_t-x_n \mid f(a)}{2}=\frac{9}{2} \iff a_1=1.73499, a_2=2.26501[/math]a_1 does not yield a solution to the problem, so only a_2 is a valid value. From here it is easy to find the (x,y) coordinates and even the equation for the tangent and normal.
...

As Dr.Peterson noted, there are indeed at least two solutions that are symmetrical about the line of symmetry of the parabola (with the point above the x-axis) and your [imath]a_1[/imath] is one of them.

Randyyy Find the point of tangent given f and the area of a triangel.

www.desmos.com

For the 5th row entry, just equate [imath]a[/imath] with 1*i, 1*j, 1*k, 1*l (see values of i, j, k, & l several rows below) to change (& see) the area indicated in the 4th row. Or you could also just slide [imath]a[/imath] from 0.10 to 4 to see how the area at row 4 varies from below 4.5 to more than 4.5

As for those two more solutions with the point below the x-axis, a little tweaking from

[imath]\frac{bh}{2}=\frac{\mid x_t-x_n \mid f(a)}{2}=\frac{9}{2}[/imath]

to

[imath]\frac{bh}{2}=\frac{\mid x_t-x_n \mid \mid f(a)\mid}{2}=\frac{9}{2}[/imath]

should do the trick.

solve \frac{1}{2}|(2a^3-12a^2+21a-8)-\frac{a^2-2}{2(a-2)}||(-a^2+4a-2)|=\frac{9}{2} - Wolfram|Alpha

Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels.

www.wolframalpha.com

I couldn't manipulate WolframAlpha to give more digits for the 3rd and 4th real number solutions though. Pretty sure pka could.

I was then also presented by a solution that used tan(x) and tan(90-x) due to the right angle formed at the top of the triangle (the point where the normal and tangent are drawn from).This piqued my interest, are there other methods/tricks to solving this problem?

Like Dr.Peterson, I am also very curious about this 2nd solution.

 

[Randyyy]

Indeed, I forgot to add that the triangel was created by the tangent, normal and of course the x-axis, The reason I disregarded my first solution is that we also know the triangel is formed in the first quadrant which I also forgot to mention it seems, my bad. It was lost during the translation as this question originally was not in english.

To the curve [math]y=2-(x-2)^2[/math] both a tangent and a normal are drawn from the same point on the curve. The tangent and the normal and the x-axis together form a triangle in the first quadrant with the area of 4.5 area units. Find the point of tangency.

This is how the question is stated,

The 2nd solution was as follows:
[math]\tan (\alpha) = \dfrac{y(a)}{b_1(a)}=y'(a)=-2a[/math][math]\tan (90 - \alpha) = \dfrac{y(a)}{b_2}=\dfrac{1}{\tan(\alpha)}[/math][math]A(a)=(b_1(a)+b_2(a))\dfrac{y(a)}{2}=\dfrac{9}{2}[/math]this gives us that:
[math]a_1 \approx -0.265, a_2 \approx -1.870[/math]If you add 2 and throw away the false solution you'll get the same point of tangency as I did with my value of 'a' being at 2.26 something.

 

[Dr.Peterson]

I had found this method, but there's a lot of work hidden in "this gives us that ..."! It's not as simple as you make it look.

But when you share a problem, you must show the entire thing. When you translate it, I recommend including the original, as well as any picture supplied, because even if no one here reads the language, we can often at least deduce that you omitted something. Not showing the whole problem violates our rules, and wastes people's time. If nothing else, it helps if we know that a problem is translated, so that we won't trust it too much.

 

[Randyyy]

I had found this method, but there's a lot of work hidden in "this gives us that ..."! It's not as simple as you make it look.

But when you share a problem, you must show the entire thing. When you translate it, I recommend including the original, as well as any picture supplied, because even if no one here reads the language, we can often at least deduce that you omitted something. Not showing the whole problem violates our rules, and wastes people's time. If nothing else, it helps if we know that a problem is translated, so that we won't trust it too much.

Yes, I agree. I left out a lot of information. Looking back, the question is not coherent at all and makes no sense the way I have phrased it. I also could have showed the graph in desmos which probably would have made it more clearer even though I had phrased it poorly.

This is exactly the way I was shown the solution, I am not 100% sure how exactly you go about solving that as we haven't expressed what b1 and b2 are so I assumed it was "obvious" and an easy calculation from then on. After seeing that you could use trigonometry I thought perhaps there was a better way of solving it than I did originally as my solution requires you to work with polynomials of higher degrees than 2 which is usually not preferred but maybe it is unavoidable this time around. It also requires you to do a lot of work. Usually trigonometry in these types of problems yields a much nicer and better solution. Better in the sense that it isn't as "brute force" as mine.

 

[Dr.Peterson]

This is exactly the way I was shown the solution, I am not 100% sure how exactly you go about solving that as we haven't expressed what b1 and b2 are so I assumed it was "obvious" and an easy calculation from then on. After seeing that you could use trigonometry I thought perhaps there was a better way of solving it than I did originally as my solution requires you to work with polynomials of higher degrees than 2 which is usually not preferred but maybe it is unavoidable this time around. It also requires you to do a lot of work. Usually trigonometry in these types of problems yields a much nicer and better solution. Better in the sense that it isn't as "brute force" as mine.

I tried this method before and got a wrong solution (though at least I did get four); I just tried again and got a 6th degree equation. I intend to try again when I have time, looking for a good trick other than the one I am already using, solving for a-2 rather than for a.

 

[Randyyy]

I tried two new and different approaches. One yielded this:

https://www.desmos.com/calculator/bocvtraodc?lang=sv-SE

Interestingly enough, if you calculate the area like you should by taking the difference between the tangent and the normals root you get an area of 3.5 which is incorrect. However, if you assume the base is the length from the origin to the tangets root you get the desired 4.5. Is this a coincidence or am I not seeing the bigger picture?.

My other attempts was trying to somehow utilize that the area could be described as -m^2/2k but this only yielded complex solutions. I think it only works had the area been contained by the curve and the tangent.

 

[Dr.Peterson]

I tried two new and different approaches. One yielded this:

https://www.desmos.com/calculator/bocvtraodc?lang=sv-SE

Interestingly enough, if you calculate the area like you should by taking the difference between the tangent and the normals root you get an area of 3.5 which is incorrect. However, if you assume the base is the length from the origin to the tangets root you get the desired 4.5. Is this a coincidence or am I not seeing the bigger picture?.

My other attempts was trying to somehow utilize that the area could be described as -m^2/2k but this only yielded complex solutions. I think it only works had the area been contained by the curve and the tangent.

I don't understand. The area of the triangle you show is about 3.6; why do you think that would be a solution at all? You seem to have found a point for which the wrong triangle has area 4.5; so it's your work that is wrong. But you haven't shown your work, so we can't see why.

 

[Randyyy]

I don't understand. The area of the triangle you show is about 3.6; why do you think that would be a solution at all? You seem to have found a point for which the wrong triangle has area 4.5; so it's your work that is wrong. But you haven't shown your work, so we can't see why.

Looking through my calculations I realize that it was a fluke. I had by accident used the points where they intersect the y-axis and gotten an incorrect value for the point of tangency which just happens to yield a point that gives 4.5 if multiplied by the height.

Perhaps trying to find an alternate solution to the two already presented are pointless. The main reason I wanted to find an alternate solution is that I am not satisfied with my original one as it felt like a lot of the steps/methods somehow could have been simplified but sometimes we don't get nice polynomials. Perhaps it is unavoidable in this problem given the restrictions.