Find the coordinates of intersection between tangents and given curve

[chwala]

Homework Statement

Find the co-ordinates of intersection between tangents that pass through the points $(-1,7)$ and the curve $y=4-x^2$.

Relevant Equations

differentiation

ooops...this was a bit tricky but anyway my approach;

...
$\dfrac{dy}{dx}=-2x$

therefore;

$\dfrac{y-7}{x+1}=-2x$

and given that, $y=4-x^2$ then;

$4-x^2-7=-2x^2-2x$

$x^2+2x-3=0$

it follows that, $(x_1,y_1)=(-3,-5)$ and $(x_2,y_2)=(1,3)$.

There may be another approach, your insight is welcome guys!.

 

[BvU]

tangents that pass through the points (1,−7)

Do they ?

$\$

 

[chwala]

Do they ?

$\$

just amended question...was a typo ...supposed to be $(-1,7)$

 

[BvU]

 

[nuuskur]

The way I understand this is: there are two lines tangent to $y:=4-x^2$ that pass through $(-1,7)$. The task is to find, where these lines intersect with $y$.

So the tangent is of the form $z-7 = k(x+1)$, where $k$ is open. We have $z=y$ such that there is exactly one point of intersection (because we want it to be tangent). So $4-x^2 = kx+ k+7$, i.e, $x^2 +kx + (k+3) = 0$. This yields the criterion $k^2 - 4(k+3)=0$ for $k$.

 

[chwala]

Aaargh @BvU Long time man! Hope you good...cheers man.

 

[WWGD]

The way I understand this is: there are two lines tangent to $y:=4-x^2$ that pass through $(-1,7)$. The task is to find, where these lines intersect with $y$.

So the tangent is of the form $z-7 = k(x+1)$, where $k$ is open. We have $z=y$ such that there is exactly one point of intersection (because we want it to be tangent). So $4-x^2 = kx+ k+7$, i.e, $x^2 +kx + (k+3) = 0$. This yields the criterion $k^2 - 4(k+3)=0$ for $k$.

K is open? Do you mean variable?

 

[DaveE]

K is open? Do you mean variable?

Yes, of course. That's just the slope of all lines that pass through the point (-1,7).

 

[WWGD]

Yes, of course. That's just the slope of all lines that pass through the point (-1,7).

Still, isn't the slope for any such line described by the derivative at the point? Let me reread the op; I may be missing something.

 

[nuuskur]

Still, isn't the slope for any such line described by the derivative at the point? Let me reread the op; I may be missing something.

One may use derivative, too. Any line passing through $(a,b)$ is of the form $y-b = k(x-a)$, where $k$ is the slope of the line.

 

[chwala]

One may use derivative, too. Any line passing through $(a,b)$ is of the form $y-b = k(x-a)$, where $k$ is the slope of the line.

...but isn't that the approach that I used? Or how different is your approach?

 

[nuuskur]

Truth be told, I don't really understand what you have done. You write out equalities but don't explain what they are or what you aim for. My crystal ball is not working, either.

 

[chwala]

Truth be told, I don't really understand what you have done. You write out equalities but don't explain what they are or what you aim for. My crystal ball is not working, either.

Kindly check my post $1$. Then let me know if you still don't understand what I did.

 

[nuuskur]

I don't have this back and forth with my students where they patronisingly tell me to re-read their work and "see if I still don't understand it".

This is not the goal. The goal is for You to understand and convincingly show that you have Understood. You may assume that I have read your attempt. It is not convincing.

$\dfrac{y-7}{x+1}=-2x$

I can infer what this means but it is not the same as you explicitly telling what it means.

and given that, $y=4-x^2$ then;
$4-x^2-7=-2x^2-2x$

Again, I understand that you substitute for $y$ in the first equality, but what is the significance of it?

In short, I'm not doubting your ability to make algebraic manipulations. I'm more interested in why you do them.

This exercise is a trivial one, so you might think it redundant to explain things. But you will carry this attitude to solutions for more complicated problems, which is undesirable.

 

[chwala]

I don't have this back and forth with my students where they patronisingly tell me to re-read their work and "see if I still don't understand it".

This is not the goal. The goal is for You to understand and convincingly show that you have Understood. You may assume that I have read your attempt. It is not convincing.

Interesting I may say...I hear you...am not a student either but I always give my students the benefit of doubt...I also can learn from them because I know that I don't know everything.My steps are clear and I fully understand the problem ...thanks for your comments though.

 

[chwala]

I don't have this back and forth with my students where they patronisingly tell me to re-read their work and "see if I still don't understand it".

This is not the goal. The goal is for You to understand and convincingly show that you have Understood. You may assume that I have read your attempt. It is not convincing.I can infer what this means but it is not the same as you explicitly telling what it means.

Again, I understand that you substitute for $y$ in the first equality, but what is the significance of it?

In short, I'm not doubting your ability to make algebraic manipulations. I'm more interested in why you do them.

This exercise is a trivial one, so you might think it redundant to explain things. But you will carry this attitude to solutions for more complicated problems, which is undesirable.

The two straight lines and the curve are intersecting at the point $(x,y)$ that's why i made the substitution.