Prove that a variable chord of ellipse

In summary, the given problem states that a variable chord of an ellipse which subtends a 90-degree angle at the center is always tangent to a concentric circle. To prove this, we use the equations of the ellipse and the circle, and homogenize them to obtain a quadratic equation. We then apply the condition of tangency and solve for the radius of the circle, which is in terms of the ellipse's parameters. However, it is also important to prove that the point of contact lies on the given chord and not on the extended line, which can be seen by drawing the figures.
  • #1
utkarshakash
Gold Member
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Homework Statement


Prove that a variable chord of ellipse which subtends 90° at the centre is always tangent to a concentric circle

Homework Equations



The Attempt at a Solution


I assume the simplest equation of ellipse to be
[itex]\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1[/itex]
and the variable chord to be
y=mx+c

Now homogenising the given line with the equation of ellipse

[itex]b^2x^2+a^2y^2-a^2b^2 \left( \dfrac{y-mx}{c} \right) ^2 = 0 \\
a^2b^2(1+m^2)=c^2(a^2+b^2) [/itex]

Now I assume the simplest equation of circle to be [itex]x^2 + y^2 = k^2[/itex]
I have to prove that y=mx+c is a tangent to given circle
Applying condition of tangency I have to prove that
[itex]k = \dfrac{c}{\sqrt{1+m^2}} [/itex]

In the RHS I substitute the value of (1+m^2) derived earlier. This way the RHS term becomes

[itex] \dfrac{ab}{\sqrt{a^2 + b^2}} [/itex]

But the LHS is k. I'm confused here!
 
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  • #2
If you pick a point O, then every line m that does not pass through O is tangent to some circle with center O. To see this, drop a perpendicular segment from O to the line m. Then use that segment as the radius of the circle.

Now take O to be the center of your ellipse and take the chord to be the one given. The chord extends to a line, so there is a concentric circle tangent to that line. But the question is whether the point of tangency lies on the original chord or whether it lies on the extended line somewhere outside the ellipse. Hint: XOY is a right triangle. Forget the ellipse and focus on the fact that you are trying to determine whether the perpendicular dropped from O to line XY lies between X and Y.
 
  • #3
Vargo said:
If you pick a point O, then every line m that does not pass through O is tangent to some circle with center O. To see this, drop a perpendicular segment from O to the line m. Then use that segment as the radius of the circle.

Now take O to be the center of your ellipse and take the chord to be the one given. The chord extends to a line, so there is a concentric circle tangent to that line. But the question is whether the point of tangency lies on the original chord or whether it lies on the extended line somewhere outside the ellipse. Hint: XOY is a right triangle. Forget the ellipse and focus on the fact that you are trying to determine whether the perpendicular dropped from O to line XY lies between X and Y.

The point of contact always lies on the chord and not on the extended line. Also what is wrong in my solution?
 
  • #4
utkarshakash said:
Now I assume the simplest equation of circle to be [itex]x^2 + y^2 = k^2[/itex]
I have to prove that y=mx+c is a tangent to given circle
Applying condition of tangency I have to prove that
[itex]k = \dfrac{c}{\sqrt{1+m^2}} [/itex]

In the RHS I substitute the value of (1+m^2) derived earlier. This way the RHS term becomes

[itex] \dfrac{ab}{\sqrt{a^2 + b^2}} [/itex]

But the LHS is k. I'm confused here!
Not sure what's bothering you. You only have to show there is some concentric circle it's tangent to. You have successfully calculated its radius.
 
  • #5
haruspex said:
Not sure what's bothering you. You only have to show there is some concentric circle it's tangent to. You have successfully calculated its radius.

Yes I later realized that I actually arrived at the answer because the RHS as well as LHS both are constant. This means the radius of the circle is the one calculated above.
 
  • #6
Would you mind explaining how you got that equation between a,b,m,c? I didn't understand that step. Does it use the fact that the chord subtends a right angle with the center? I can't see how you got there, but it looks like a neat trick.
 
  • #7
Vargo said:
Would you mind explaining how you got that equation between a,b,m,c? I didn't understand that step. Does it use the fact that the chord subtends a right angle with the center? I can't see how you got there, but it looks like a neat trick.

Yes Of course. See my question. It says that the chord subtends a right angle at the centre. So I used the concept of homogenisation to arrive at that equation.
 
  • #8
I see. You homogenize the two equations to get a quadratic form whose zero set is a pair of orthogonal lines. Therefore the trace of the form is zero giving you the equation. Cool trick.

I am still not clear on your proof though. You have a nice calculation of the radius of the tangent circle in terms of the parameters of the ellipse a and b. But have you proved that the point of contact is on the chord rather than the extended line y=mx+b? Somehow, you must address this explicitly because that is the crux of what you are being asked to prove. (Otherwise the subtended angle would be irrelevant and the statement would be trivial).
 
  • #9
Vargo said:
I see. You homogenize the two equations to get a quadratic form whose zero set is a pair of orthogonal lines. Therefore the trace of the form is zero giving you the equation. Cool trick.

I am still not clear on your proof though. You have a nice calculation of the radius of the tangent circle in terms of the parameters of the ellipse a and b. But have you proved that the point of contact is on the chord rather than the extended line y=mx+b? Somehow, you must address this explicitly because that is the crux of what you are being asked to prove. (Otherwise the subtended angle would be irrelevant and the statement would be trivial).

You can try out by drawing the figures. You will see that the point of contact always lies on the chord and not on the extended line if the circle is concentric.
 

FAQ: Prove that a variable chord of ellipse

What is an ellipse?

An ellipse is a geometric shape that resembles a flattened circle. It has two focal points, or foci, and is defined by the distance between these foci and the points on its perimeter.

What is a variable chord of an ellipse?

A variable chord of an ellipse is a line segment that connects two points on the ellipse's perimeter and can vary in length and position.

How do you prove that a variable chord of an ellipse?

To prove that a variable chord of an ellipse, we use the property that the sum of the distances from any point on the ellipse to the foci is constant. We can set up an equation using this property and the Pythagorean theorem to show that the length of the chord is dependent on the distance from the foci to the point on the ellipse.

What is the importance of proving the variable chord of an ellipse?

Proving the variable chord of an ellipse is important in understanding the geometric properties of an ellipse and its relationship to its foci. It also allows for the calculation of the length of a chord given its distance from the foci, which can be useful in various mathematical and scientific applications.

Are there any real-world applications of the variable chord of an ellipse?

Yes, the variable chord of an ellipse has many real-world applications, including in optics, architecture, and engineering. For example, in optics, the shape of an ellipse is used in the design of lenses and mirrors to manipulate light. In architecture, ellipses are often used in the design of domes and arches. In engineering, ellipses can be used to model the orbits of planets and satellites.

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