Finding tangent point between a line and a hyperbola

Sap67

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May 26, 2020
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Hi, so I have the following exercice as a homework, it's in french so I'll do my best to translate it.


We got \(\displaystyle y = f(x) = \frac{3x + ax + b}{x^2+1}\)
where a and b are real numbers.
Determine a and b the curve f(x) is tangent to the line \(\displaystyle y = 4x + 3\) on the point \(\displaystyle I(0;3) \).


Before I explain what I tried so far, I have to say, I haven't really started school yet. We only got some homework by some new teachers, and what this teacher has been sending us is slightly more difficult from what we learned last year (Which is much, mostly because of the pandemic, and some other issues my country had to deal with which caused schools to close a lot).

So I tried some things we would do last year like :
- Combining both equations then doing the quadratic equation in order to get this : \(\displaystyle \delta = x^2 + 4ac \)
Then comparing it to 0 so there's only one possibility for x.
But that doesn't work here since the degree of the equation I get is 3 and I found no way of simplifying it. And we haven't learned how to deal with third degree equations.

-Using the following formula : \(\displaystyle y = f'(a) (x-a) + f(a) \) .

- Combining both of the methods above in different orders.

-Replacing x and y in the curve equation by those given by the I point, also combined with the methods above.

But none of the things I tried worked out.

This is what I got after typing the equations I got on desmo :
While they are intersecting on 0;3, they are nowhere near tangent
 

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Jomo

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You have two unknowns, namely a and b. So you need two pieces of information to solve for them.

One piece is that the two equations meet at (0,3). That gives us that f(0) = 3. This tells us the value for b. What is it?

The slope of the line and the slope of the tangent line of f(x) at (0,3) must be equal. That gives you another equation. What is it? Now solve for a.
 

JeffM

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Sep 14, 2012
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My suggestion is very similar to Jomo's, but I might do it the other way round by starting with the line and cleaning up the notation.

First, I hate using the same variable for two different things. That seems to me to invite conceptual confusion.

So we know

\(\displaystyle y(x) = 4x + 3 \implies y'(x) = 4.\)

\(\displaystyle z(x) = \dfrac{3x + ax + b}{x^2 + 1} \implies z'(x) = \dfrac{(3 + a)(x^2 + 1) - (3 + a)x(2x)}{(x^2 + 1)^2} =\)

\(\displaystyle \dfrac{3x^2 + 3 + ax^2 + a - 6x^2 - 2ax^2}{(x + 1)^2} = \dfrac{3 + a - x^2(3a + 1)}{((x^2 + 1)^2}.\)

\(\displaystyle y(0) = z(0) \text { and } y'(0) = z'(0).\)

Why?

Now you are just in the same place as Jomo. My way walks; his way runs.
 

Sap67

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Joined
May 26, 2020
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You have two unknowns, namely a and b. So you need two pieces of information to solve for them.

One piece is that the two equations meet at (0,3). That gives us that f(0) = 3. This tells us the value for b. What is it?

The slope of the line and the slope of the tangent line of f(x) at (0,3) must be equal. That gives you another equation. What is it? Now solve for a.
First of all, thank you for taking time out of your day to help me, I really appreciate it.
So what I did is :
First of all, I got b = 3 by doing f(0) = 3, which I had already done before but used the answer with the wrong methods

Then I got the slupe for the curve which was : \(\displaystyle \frac{(a-2)x^2+a}{(x^2+1)^2} \) which I compared to 4, which is the line's slope and I got the following :
\(\displaystyle 4x^4+ (10-a)x^2 + 4-a = 0 \) but then from here I don't know what to do because, everything I try I seem to hit a dead end
Hitting dead ends when trying to use the formula, and same thing when I try to compare this equation with any other one.

My suggestion is very similar to Jomo's, but I might do it the other way round by starting with the line and cleaning up the notation.

First, I hate using the same variable for two different things. That seems to me to invite conceptual confusion.

So we know

\(\displaystyle y(x) = 4x + 3 \implies y'(x) = 4.\)

\(\displaystyle z(x) = \dfrac{3x + ax + b}{x^2 + 1} \implies z'(x) = \dfrac{(3 + a)(x^2 + 1) - (3 + a)x(2x)}{(x^2 + 1)^2} =\)

\(\displaystyle \dfrac{3x^2 + 3 + ax^2 + a - 6x^2 - 2ax^2}{(x + 1)^2} = \dfrac{3 + a - x^2(3a + 1)}{((x^2 + 1)^2}.\)

\(\displaystyle y(0) = z(0) \text { and } y'(0) = z'(0).\)

Why?

Now you are just in the same place as Jomo. My way walks; his way runs.
I am extremely sorry, I made a mistake while copying the equation, it should be (3x^2 +ax +b)/(x^2+1)
Sorry I wasted your time on that :/ But I tried following what you did, I get it but I don't get what to do after since every value I got by trying the quadratic equation, it was never right.
 

JeffM

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Sep 14, 2012
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5,231
You found b = 3 by equating y(0) and z(0). Why can you do that? Exactly what math did you do to come up with b = 3? Once you know that however, you can and should restate z(x) as

\(\displaystyle z(x) = \dfrac{3x^2 + ax + 3}{x^2 + 1} \implies z'(x) = \dfrac{(6x + a)(x^2 + 1) - (3x^2 + ax + 3)(2x)}{(x^2 + 1)^2} =\\ \dfrac{6x^3 + 6x + ax^2 + a - 6x^3 - 2ax^2 - 6x}{(x ^2 + 1)^2} = \dfrac{a(1 - x^2)}{(x^2 + 1)^2}.\)
So check your differentiation.

Then you went completely off the rails. Where did the quartic come from? You are not trying to solve for x. z'(x) is a function rather than an unknown. You can equate y'(0) = z'(0). Why?

Assuming my differentiation is correct that requires solving for a in

\(\displaystyle 4 = \dfrac{a(1 - 0^2)}{(0^2 + 1)}\)
 
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