Distance from a point on an ellipse to the axis

[JimEd]

Hi,

Concerning only the upper right quadrant of an ellipse...

I know the distance from the center of the ellipse to the top of the ellipse, (semi-minor axis "b"), is 1000.
I know the distance from the center of the ellipse to the side of the ellipse, (semi-major axis "a") is 1732.

At the point where the slope of the ellipse is at 45 degrees, if you drew a line from that point, perpendicular to that 45 degree slope, inward to where it intersects "a", what would be the distance?

Thanks!

(edited to clarify that I meant the slope)

 

[Mark44]

Hi,

Concerning only the upper right quadrant of an ellipse...

I know the distance from the center of the ellipse to the top of the ellipse, (semi-minor axis "b"), is 1000.
I know the distance from the center of the ellipse to the side of the ellipse, (semi-major axis "a") is 1732.

At the point where the arc of the ellipse is at 45 degrees, if you drew a line from that point, perpendicular to that 45 degree angle, inward to where it intersects "a", what would be the distance?

Thanks!

I don't understand what you're asking. What does "arc of the ellipse is at 45°" mean? Do you mean the point on the ellipse that is intersected by the line y = x?

And we don't talk about lines being perpendicular to angles - a line can be perpendicular to another line, but not an angle.

 

[JimEd]

I don't understand what you're asking. What does "arc of the ellipse is at 45°" mean? Do you mean the point on the ellipse that is intersected by the line y = x?

And we don't talk about lines being perpendicular to angles - a line can be perpendicular to another line, but not an angle.

Sorry, let's try it this way...

Not the point intersected by line y=x. The point where the actual slope of the ellipse is at 45 degrees.

And perpendicular to that slope.

(edited the original question to include this clarification)

 

[JimEd]

I would also like to understand how it's calculated, in addition to having the answer, if that's possible. Thank you.

 

[Norwegian]

Hi, I understand your post as asking about the y-coordinate of the first quadrant point where the ellipse has slope -1.

This can be done straight forwardly by writing the equation of the ellipse, then by implicit derivation, or otherwise, find an expression for dy/dx, and then solve the equations. My answer was suspiciously close a round number, which I guess reflects the fact that the major axis is close to √3 times the minor axis, 1000, and which also make me believe there is a more conceptual way to solve this.

edit: and I see now that you are asking about the distance along the perpendicular to the tangent, so my y-coordinate, 500+change, must be multiplied by √2 to answer your question.

 

[JimEd]

Hi, I understand your post as asking about the y-coordinate of the first quadrant point where the ellipse has slope -1.

This can be done straight forwardly by writing the equation of the ellipse, then by implicit derivation, or otherwise, find an expression for dy/dx, and then solve the equations. My answer was suspiciously close a round number, which I guess reflects the fact that the major axis is close to √3 times the minor axis, 1000, and which also make me believe there is a more conceptual way to solve this.

Thanks for working on my question. Most of what you are saying is way over my head, is there a simpler way that you or someone else could walk me through?

What answer did you get?

Thanks.

 

[Mark44]

Thanks for working on my question. Most of what you are saying is way over my head, is there a simpler way that you or someone else could walk me through?

I don't think there is. To answer the question you're asking, you need to understand these concepts:

• finding the equation of an ellipse
• finding the slope of a tangent line to a curve
• finding the equation of a line, given its slope and a point on it
• finding the point of intersection of two lines

If you don't have a basic understanding of these concepts, I don't think you would be able to understand an explanation.

What answer did you get?

Thanks.

 

[JimEd]

I don't think there is. To answer the question you're asking, you need to understand these concepts:

• finding the equation of an ellipse
• finding the slope of a tangent line to a curve
• finding the equation of a line, given its slope and a point on it
• finding the point of intersection of two lines

If you don't have a basic understanding of these concepts, I don't think you would be able to understand an explanation.

I am willing to try!

 

[JimEd]

I don't think there is. To answer the question you're asking, you need to understand these concepts:

• finding the equation of an ellipse
• finding the slope of a tangent line to a curve
• finding the equation of a line, given its slope and a point on it
• finding the point of intersection of two lines

If you don't have a basic understanding of these concepts, I don't think you would be able to understand an explanation.

And this is not homework (I'm almost 50 years old). I just want to (1) know the answer and (2) understand how it was arrived at, to the best of my ability.

 

[JimEd]

Hello? Anyone?...

 

[JimEd]

Hi, I understand your post as asking about the y-coordinate of the first quadrant point where the ellipse has slope -1.

This can be done straight forwardly by writing the equation of the ellipse, then by implicit derivation, or otherwise, find an expression for dy/dx, and then solve the equations. My answer was suspiciously close a round number, which I guess reflects the fact that the major axis is close to √3 times the minor axis, 1000, and which also make me believe there is a more conceptual way to solve this.

edit: and I see now that you are asking about the distance along the perpendicular to the tangent, so my y-coordinate, 500+change, must be multiplied by √2 to answer your question.

Hi,

I don't see your answer or your y-coordinate... am I missing something?

Thanks

 

[Norwegian]

Yes,
y-coordinate ≈ 500.01
answer ≈ 500.01*√2
Does that sound reasonable? I havnt double checked.

 

[JimEd]

Yes,
y-coordinate ≈ 500.01
answer ≈ 500.01*√2
Does that sound reasonable? I havnt double checked.

Thanks so much for replying to me.

Your answer does sound reasonable to me based on a drawing of it, but I really wouldn't know math-wise.

So, the 500.01 is basically 1/2 b? Is that why it is 500?

 

[JimEd]

Do you think the 45 degree point will always be at .5b for all ellipses? That would make sense, right?

And if that is true, then .5b*sqr2 will be the answer to this problem for all ellipses, would that be correct?

 

[Mark44]

Do you think the 45 degree point will always be at .5b for all ellipses? That would make sense, right?

I don't think so.

If we look at a circle, which is a sort of special case of an ellipse (with a = b), and the point at which the tangent has a slope of -1 (what you're calling the 45° point), the perpendicular to the tangent line goes to the center of the circle.

As we broaden the circle to make an ellipse, that perpendicular is going to hit farther out on the x-axis. If you had an ellipse with a larger semiaxis than your ellipse, you get a larger value for the distance from the center to the intersection point on the x-axis.

And if that is true, then .5b*sqr2 will be the answer to this problem for all ellipses, would that be correct?

 

[JimEd]

OK, I see what you are saying.

So then, if we draw a horizontal line from the 45 degree point over to the y-axis and read that value, then multiply that value times sqr2, would that be the answer to this problem for all ellipses? (it works on the circle I drew, and the ellipse)

How would you state that as a mathematical formula, preferably as a proportion/ratio of a and/or b (I am trying to keep it free of any specific coordinate system)?

 

[Norwegian]

Do you think the 45 degree point will always be at .5b for all ellipses? That would make sense, right?

No. And it is not even true for your ellipse. As I indicated above, the y-coordinate of the 45 degree point is slightly more than b/2.

So then, if we draw a horizontal line from the 45 degree point over to the y-axis and read that value, then multiply that value times sqr2, would that be the answer to this problem for all ellipses?

Yes. But that has nothing to do with ellipses, it is true for any first quadrant point on a 45 degree angled line, and is the same statement as Pythagoras on a right angled equilateral triangle.

 

[Mark44]

OK, I see what you are saying.

So then, if we draw a horizontal line from the 45 degree point over to the y-axis and read that value, then multiply that value times sqr2, would that be the answer to this problem for all ellipses? (it works on the circle I drew, and the ellipse)

How would you state that as a mathematical formula, preferably as a proportion/ratio of a and b (I am trying to keep it free of any specific coordinate system)?

Here's some math.

Let's assume that the equation of the ellipse is
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

It turns out that the derivative, dy/dx, for this ellipse is
dy/dx = -(b²x)/(a²y)

The derivative gives us the slope of the tangent line at any point.

At the point you're interested in, the slope is -1, so we set the derivative to -1 and solve.

-1 = -(b²x)/(a²y)
==> 1 = (b²x)/(a²y)
==> a²y = b²x
==> y = (b²/a²)x

The last equation gives the y value on the ellipse at which the slope of the tangent line is -1.

For a circle, a = b, so the point of interest is on the line y = x. For an ellipse, you need to put in the values of a and b and use the fact that this point is on the ellipse, which means that it satisfies the ellipse's equation.

You now have two equations in the two unknowns x and y, so you can find the point of interest.

After that you can run a horizontal line from this point over to the y axis, or you can figure out where a line with slope 1 hits the x-axis.

 

[JimEd]

Here's some math.

Let's assume that the equation of the ellipse is
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

It turns out that the derivative, dy/dx, for this ellipse is
dy/dx = -(b²x)/(a²y)

The derivative gives us the slope of the tangent line at any point.

At the point you're interested in, the slope is -1, so we set the derivative to -1 and solve.

-1 = -(b²x)/(a²y)
==> 1 = (b²x)/(a²y)
==> a²y = b²x
==> y = (b²/a²)x

The last equation gives the y value on the ellipse at which the slope of the tangent line is -1.

For a circle, a = b, so the point of interest is on the line y = x. For an ellipse, you need to put in the values of a and b and use the fact that this point is on the ellipse, which means that it satisfies the ellipse's equation.

You now have two equations in the two unknowns x and y, so you can find the point of interest.

After that you can run a horizontal line from this point over to the y axis, or you can figure out where a line with slope 1 hits the x-axis.

Thanks for this.

How do I find y using:
y = (b²/a²)x
without knowing x?

And then how would I find x, plug y back into the same formula?

You now have two equations in the two unknowns x and y, so you can find the point of interest.

Which two equations? The one above, and... ?

 

[JimEd]

In other words, y=(b2/a2)x basically says y is a certain fraction of x, but to get the actual number for y, I would need to know x.

Or, if I know y, I can calculate x.

But I need to know one of the two in order to get the other, right? Or am I missing some math trick that accomplishes this?

 

[JimEd]

Can anyone offer any help on this?

What is the formula to find y, knowing just a and b?

 

[JimEd]

Here's some math

Hi Mark,

Can you offer any more help on finding y?

Thank you.

 

[Mark44]

Thanks for this.

How do I find y using:
y = (b²/a²)x
without knowing x?

And then how would I find x, plug y back into the same formula?



Which two equations? The one above, and... ?

The one for the ellipse - x²/a² + y²/b² = 1. You can replace y in this equation by (b²/a²)x, and you'll have one equation in x alone, which you can solve for x.

 

[JimEd]

The one for the ellipse - x²/a² + y²/b² = 1. You can replace y in this equation by (b²/a²)x, and you'll have one equation in x alone, which you can solve for x.

Mark,

I'm sorry but I have spent the last few hours working on this and I'm just not getting it.

x2/a2 + y2/b2 = 1, replace y with (b2/a2)x =

x2/a2+(((b2/a2)x)((b2/a2)x))/b2=1

I'm sure that simplifies down to something, but damned if I know. And even then it gives me x, and I want y.

Isn't there a simplified version of this formula for "y="? I've just about had it with this. I'm frustrated as hell.

 

[JimEd]

The one for the ellipse - x²/a² + y²/b² = 1. You can replace y in this equation by (b²/a²)x, and you'll have one equation in x alone, which you can solve for x.

I did come up with a formula for the length I am looking for (from the 45 degree slope). It's not as simple as Norwegian's y*sqr2 for my purposes though, and I would still have to find x.

I just need a simple "y=" formula, that uses only a and b. Then I multiply that times sqr2 and I am done. Do you know what that would be? Is it possible?

 

[JimEd]

Anyone?

What is the simplest formula to find y for an ellipse, given only a and b?

y= ...?

This is not homework. I just need the answer.

Thank you.

 

[Mark44]

Assuming that your ellipse is centered at the origin, its equation is
x²/a² + y²/b² = 1

Solving for y, we get
y²/b² = 1 - x²/a²
==> y² = b²(1 - x²/a²)
==> y = ± √(b²(1 - x²/a²))

For points above the x-axis, use the positive square root; for points below the x-axis, use the negative square root.

 

[JimEd]

Thank you again for your help, everyone.