Tangent line through point not on function

[gGo]

We want to find all the points on the graph of [MATH]y=\frac 2x + 3x[/MATH] where the tangent line at that point passes through [MATH](6, 17)[/MATH]. Notice that point is not on the graph of y.

The slope of the tangent line is given by $$y'= -\frac{2}{x^2} + 3$$. But we don't know what $$x$$ is.

A tangent line with that slope, through that point is

$$y = 17 + \left(-\frac{2}{x^2} + 3\right)(x-6)$$.​

Working back from one of the answers, (-6, -55/3), and from a partial solution we were given, I think we can calculate the slope of the tangent at $$x=6$$. That is, the slope of one tangent is
$$m = -\frac{2}{6^2}+3 = 53/18$$.

Now that we know the slope, we can find the points on the curve that have that slope by finding where the derivative equals it
$$-\frac{2}{x^2} + 3 = 53/18$$$$x = -6, 6$$
x=6 is clearly extraneous.
x=-6 works (I checked by graphing and other methods).

The other given solution, $$(2, 7)$$ also works, but does not give me any clues about how to find it.

Are we allowed to substitute x=6 into slope? If so, why? How do I finish this problem?

Thanks.

 

[apple2357]

I started with let a point P on the curve be (a, (2/a)+3a)
Find the equation of the tangent at this point ( which lies on the curve) and then solve for a when x=6, y=17
I found a is -6 and 2. Its a bit messy though.
Not quite sure how you got -55/3 above

 

[JeffM]

A line is defined by two points, correct?

One of the defining points on each the desired lines is (6, 17).

What can we say about the other defining point, which I shall call (p, q) to avoid confusion with general x and y.

It must be on the curve y = (2/x) + 3x.

[MATH]\therefore q = \dfrac{2}{p} + 3p = \dfrac{3p^2 + 2}{p}.[/MATH]
Hmm, one equation and two unknowns. What else do we know that might help us?

We know that the slope of the line is

[MATH]m = \dfrac{q - 17}{p - 6}.[/MATH]
Great, just great. Now we have two equations and three unknowns.

Is there a third clue?

There is. We know that this line is tangent to the curve at (p, q) and that the slope of the tangent line at a point is equal to the first derivative at that point.

[MATH]\dfrac{dy}{dx} = - \dfrac{2}{x^2} + 3 \implies m = - \dfrac{2}{p^2} + 3 \implies \dfrac{q - 17}{p - 6} = - \dfrac{2}{p^2} + 3 = \dfrac{3p^2 - 2}{p^2} \implies[/MATH]
[MATH]\therefore p^2q - 17p^2 = 3p^3 - 18p^2 - 2p + 12 \implies \dfrac{p^2(3p^2 + 2)}{p} - 17p^2 = 3p^3 - 18p^2 - 2p + 12 \implies[/MATH]
[MATH]3p^3 + 2p - 17p^2 = 3p^3 - 18p^2 - 2p + 12 \implies p^2 + 4p - 12 = 0 \implies[/MATH]
[MATH](p + 6)(p - 2) = 0 \implies p = - 6 \text { or } p = 2.[/MATH]
Now what?

MORAL OF THIS PROBLEM: You need as many clues to turn into equations as you have unknowns. Here the slopes of the lines and the coordinates of the points on the curve were all unknown so you must find three clues.

EDIT: This is just a more detailed way to express apple’s answer.

 

[gGo]

How are you finding the equation of the tangent?

The point is $$(a, 2/a+3a)$$ and the derivative is $$y = -2/x^2 + 3$$. Then the slope is $$m = -2/a^2 + 3$$, then the tangent line is

$$y - \left(\frac 2a + 3a\right) = \left(-\frac{2}{a^2} + 3\right)\left(x-a\right)$$​

$$y = \frac{3a^2-2}{a^2}x + \frac 4a$$​

If that's right, then what? When x=6, the tangent line is

$$y = \frac{3a^2-2}{a^2}(6) + \frac 4a$$​

And I'm not sure what to do from there.

 

[gGo]

A line is defined by two points, correct?

One of the defining points on each the desired lines is (6, 17).

What can we say about the other defining point, which I shall call (p, q) to avoid confusion with general x and y.

It must be on the curve y = (2/x) + 3x.

[MATH]\therefore q = \dfrac{2}{p} + 3p = \dfrac{3p^2 + 2}{p}.[/MATH]
Hmm, one equation and two unknowns. What else do we know that might help us?

We know that the slope of the line is

[MATH]m = \dfrac{q - 17}{p - 6}.[/MATH]
Great, just great. Now we have two equations and three unknowns.

Is there a third clue?

There is. We know that this line is tangent to the curve at (p, q) and that the slope of the tangent line at a point is equal to the first derivative at that point.

[MATH]\dfrac{dy}{dx} = - \dfrac{2}{x^2} + 3 \implies m = - \dfrac{2}{p^2} + 3 \implies \dfrac{q - 17}{p - 6} = - \dfrac{2}{p^2} + 3 = \dfrac{3p^2 - 2}{p^2} \implies[/MATH]
[MATH]\therefore p^2q - 17p^2 = 3p^3 - 18p^2 - 2p + 12 \implies \dfrac{p^2(3p^2 + 2)}{p} - 17p^2 = 3p^3 - 18p^2 - 2p + 12 \implies[/MATH]
[MATH]3p^3 + 2p - 17p^2 = 3p^3 - 18p^2 - 2p + 12 \implies p^2 + 4p - 12 = 0 \implies[/MATH]
[MATH](p + 6)(p - 2) = 0 \implies p = - 6 \text { or } p = 2.[/MATH]
Now what?

MORAL OF THIS PROBLEM: You need as many clues to turn into equations as you have unknowns. Here the slopes of the lines and the coordinates of the points on the curve were all unknown so you must find three clues.

EDIT: This is just a more detailed way to express apple’s answer.

Thanks. This is a great answer.
Next we will find q.

 

[apple2357]

How are you finding the equation of the tangent?

The point is $$(a, 2/a+3a)$$ and the derivative is $$y = -2/x^2 + 3$$. Then the slope is $$m = -2/a^2 + 3$$, then the tangent line is

$$y - \left(\frac 2a + 3a\right) = \left(-\frac{2}{a^2} + 3\right)\left(x-a\right)$$​

$$y = \frac{3a^2-2}{a^2}x + \frac 4a$$​

If that's right, then what? When x=6, the tangent line is

$$y = \frac{3a^2-2}{a^2}(6) + \frac 4a$$​

And I'm not sure what to do from there.

When x =6, y=17 and you can solve for a
Another approach might be to use disriminant.

 

[gGo]

When x =6, y=17 and you can solve for a
Another approach might be to use disriminant.

Oh yeah, duh! (and TY)

 

[JeffM]

Thanks. This is a great answer.
Next we will find q.

Exactly. And we already have q in terms in terms of p

[MATH]q = \dfrac{3p^2 + 2}{p}.[/MATH]
So it’s just plug and chug. Nothing fancy required.