Thread: Find the intersection of a curve and a normal if the normal passes by a given point

Hi!

I'm having trouble with the following exercise and would like some help.

I need to find the intersection point of a curve and a normal to the curve knowining only the curve equation and a coordinate the normal passes by (which is not the intersection between the curve and the normal)

The curve equation is and the normal passes by the point

Thanks!

Originally Posted by Kevindqc

I'm having trouble with the following exercise and would like some help. I need to find the intersection point of a curve and a normal to the curve knowining only the curve equation and a coordinate the normal passes by (which is not the intersection between the curve and the normal)
The curve equation is and the normal passes by the point

A normal to the curve is perpendicular to the tangent.

Suppose that is the point we seek.
Then .

But the slope of the normal there is . WHY?

Hi Kevindqc!

You've got a difficult problem there.
If you solve it analytically, you'll run into a third degree polynomial that does not have a neat solution.

Is it possible you are supposed to solve it using graphical methods?

Originally Posted by Plato

A normal to the curve is perpendicular to the tangent.

Suppose that is the point we seek.
Then .

But the slope of the normal there is . WHY?

Hmm well I get that the slope of the normal is since the derivative is , the normal slope multiplied by the slope must be -1 to be perpendicular.

So the normal slope is , which is also , the quotient of the y and x deltas.

But I still don't know how to get a and b :'(

Originally Posted by ILikeSerena

Hi Kevindqc!

You've got a difficult problem there.
If you solve it analytically, you'll run into a third degree polynomial that does not have a neat solution.

Is it possible you are supposed to solve it using graphical methods?

Hello Serena liker,

Well there's a graph to illustrate the problem, but I'm not supposed to use it, no.
The exercice does have a star, which means it's a hard problem.

I suppose to teacher will explain it in class, but I had surgery and I'm gonna miss a couple weeks of school

Originally Posted by Kevindqc

Well there's a graph to illustrate the problem, but I'm not supposed to use it, no.
The exercice does have a star, which means it's a hard problem.

I suppose to teacher will explain it in class, but I had surgery and I'm gonna miss a couple weeks of school

Are you then supposed to use a calculator that can solve sets of equations?
One that would give you an approximation of the answer?

You have the 2 equations that Plato provided.
Do you know how to solve such a set of equations?
Basically you need to substitute the first in the second, that is, replace the occurrences of b in the second equation, by the b given by the first equation.

Originally Posted by Kevindqc

I suppose to teacher will explain it in class, but I had surgery and I'm gonna miss a couple weeks of school

You can use the web. See here.

Originally Posted by ILikeSerena

Are you then supposed to use a calculator that can solve sets of equations?
One that would give you an approximation of the answer?

You have the 2 equations that Plato provided.
Do you know how to solve such a set of equations?
Basically you need to substitute the first in the second, that is, replace the occurrences of b in the second equation, by the b given by the first equation.

Heya,

Oh, I see it now!
Yes, I have the mandatory expensive calculator, a ti-nspire cx cas :/

It's the first time I see it taking a couple seconds to solve something, but I got a = 3.78377 and b = 2.81817, the correct answer!

My thanks to you and Plato!

Arg sorry for being annoying, but the next question asks for the coordinate (7; 2) but when I try to solve using the above, I get 3 possible answers...? There's supposed to be 2 answers, but none of the 3 Given by wolframalpha are good :/

The answers of Wolfram are good.
You can verify if you try to draw it.

Or you can take a look at this graph by Wolfram.

Originally Posted by ILikeSerena

The answers of Wolfram are good.
You can verify if you try to draw it.

Or you can take a look at this graph by Wolfram.

Maybe the solutions are wrong, but how is it possible to have 3 intersections? 1 or 2, yeah, but 3? It's a parabola, it can't be, no?

Originally Posted by Kevindqc

Maybe the solutions are wrong, but how is it possible to have 3 intersections? 1 or 2, yeah, but 3? It's a parabola, it can't be, no?

Consider a point in the middle of the parabola.
It would have a normal to the left, a symmetric normal to the right.... and a normal to the top!

Point is, you have an equation of the third degree.
It can have 1, 2, or 3 solutions.

In your case, 2 of the solutions are close together, near x=1.

ok, there can be more than one result if the point is inside the parabola.

But that's not the case here, if you look at the graph, you can see that the point (7, 2) is outside to the right, so I still don't get why there can be more than 2 intersections!


But anyway, I'm a complete fool. I misread. The point (7, 2) is touching a tangent, not a normal

Never mind I get it now

Originally Posted by Kevindqc

ok, there can be more than one result if the point is inside the parabola.

But that's not the case here, if you look at the graph, you can see that the point (7, 2) is outside to the right, so I still don't get why there can be more than 2 intersections!

I just mentioned the point inside the parabola to make my point.
It also applies outside the parabola - just not everywhere.
Usually there will be either 1 point or 3 points.
It will be rare if there are 2.

But anyway, I'm a complete fool. I misread. The point (7, 2) is touching a tangent, not a normal

Aha! That explains it!