# FP3 Coordinate systems help ! :/

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I don't know where to start for part c, the mark scheme suggests substitution y=mx+c in the ellipse equation but i dont understand why?!

Q http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf

MS http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf

Q http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf

MS http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf

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So what it's saying is that if you fix a particular gradient, m, and then draw all the chords with that gradient, the midpoints of the chords will lie on a straight line.

So one approach is going to be:

Take an arbitrary line with gradient m (i.e. y = mx + c).

Find where this arbitrary line intersects the ellipse.

If it intersects the ellipse at 2 points, find the midpoint.

Show this midpoint always lies on a straight line. (i.e. the intersection points will be a function of c, and therefore so will the midpoint. You want to show that there's a particular straight line that the midpoint always lies on as you vary c).

So one approach is going to be:

Take an arbitrary line with gradient m (i.e. y = mx + c).

Find where this arbitrary line intersects the ellipse.

If it intersects the ellipse at 2 points, find the midpoint.

Show this midpoint always lies on a straight line. (i.e. the intersection points will be a function of c, and therefore so will the midpoint. You want to show that there's a particular straight line that the midpoint always lies on as you vary c).

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(Original post by

So what it's saying is that if you fix a particular gradient, m, and then draw all the chords with that gradient, the midpoints of the chords will lie on a straight line.

So one approach is going to be:

Take an arbitrary line with gradient m (i.e. y = mx + c).

Find where this arbitrary line intersects the ellipse.

If it intersects the ellipse at 2 points, find the midpoint.

Show this midpoint always lies on a straight line. (i.e. the intersection points will be a function of c, and therefore so will the midpoint. You want to show that there's a particular straight line that the midpoint always lies on as you vary c).

**DFranklin**)So what it's saying is that if you fix a particular gradient, m, and then draw all the chords with that gradient, the midpoints of the chords will lie on a straight line.

So one approach is going to be:

Take an arbitrary line with gradient m (i.e. y = mx + c).

Find where this arbitrary line intersects the ellipse.

If it intersects the ellipse at 2 points, find the midpoint.

Show this midpoint always lies on a straight line. (i.e. the intersection points will be a function of c, and therefore so will the midpoint. You want to show that there's a particular straight line that the midpoint always lies on as you vary c).

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I dont know how to find c- ive tried subbing x into the quadratic equation but my value isn't right

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I dont know how to find c- ive tried subbing x into the quadratic equation but my value isn't right

**FHL123**)I dont know how to find c- ive tried subbing x into the quadratic equation but my value isn't right

Take a generic line y = mx + c.

Find the two points where it intersects the ellipse.

Find the midpoint (x(c), y(c)) of those points. (I've written this as x(c), y(c) because it will depend on c).

Show that we can find constants A, B, C s.t.

A x(C) + B y(c) = C for all choices of c.

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You don't "find" c.

Take a generic line y = mx + c.

Find the two points where it intersects the ellipse.

Find the midpoint (x(c), y(c)) of those points. (I've written this as x(c), y(c) because it will depend on c).

Show that we can find constants A, B, C s.t.

A x(C) + B y(c) = C for all choices of c.

**DFranklin**)You don't "find" c.

Take a generic line y = mx + c.

Find the two points where it intersects the ellipse.

Find the midpoint (x(c), y(c)) of those points. (I've written this as x(c), y(c) because it will depend on c).

Show that we can find constants A, B, C s.t.

A x(C) + B y(c) = C for all choices of c.

I do not know the marking scheme, and there is another way for way to do this.

It is to combine, straight line equation form of y - a = M (x -b), mid point, algebra and recognising patterns (i.e. mathematical deductions).

So, lets start. Say, two such point A (p, q) and B (r, s).

Equation for elipse is x^2/a2 + y^2/b^2 = 1. Midpoint of AB, say D ( (p+r)/2, (q+s)/2) ).

Substitute coordinates of A and B on elipse's equation, because A and B lie onelipse itself.

(p^2/a^2 + q^2/b^2) = 1 (E.1)

(r^2/a^2) + (s^2/b^2) = 1 (E.2)

Now because midpoint of AB is ( (p+r)/2, (q+s)/2) ) and (q-s) / (p-r) is gradient of AB (gradient is m, given),we see that if we deduct E.2 from E.1 and rearrange, we can get the double value of midpoint and the gradient of AB.

So, ( (p-r)(p+r) / (a^2)) + ( (q-s)(q+s) / (b^2) ) = 0

Rearrange to form the straight line equation from y - a = M (x -b)

(q-s)(q+s) = -(b^2 / a^2) [((p-r)(p+r)]

(q+s) /2 = -(b^2 / a^2)(1/m) [(p+r) / 2]

So the midpoint of chord AB, D satisfy the above equation.

Any midpoint of two points on the elipse (i.e. midpoint of chord) will also satisfy the above equation as long as -(b^2 / a^2)(1/m) remains constant because points A, B always change (so do the coordinates), and for -(b^2 / a^2)(1/m) to remain constant, m has to remain same for any two points (i.e for any chords) and thus chords have to be parallel.

The equation is in the form of (y - 0) = M(x - 0), where M = -( b^2 / (a^2 x m) ) and this is one form of straight line equation.

Therefore, midpoints of parallel chords will always lie on straight line, and the straight line through the midpoints of parallel chords will always go through the origin. i.e. intercept is always 0, and the gradient of the straight line through the midpoints of parallel chords is -( b^2 / (a^2 x m) ).

This can easily verifed using the midpoints of chords parllel either to major axis or minor axis (which are special cases of chords), where origin itself is midpoint and satifies the above equation.

In the case of chords are parallel to x axis, m = 0, which is y axis itself, and in the case of

chords are parallel to y axis, m approaches infinity, which is x axis itself.

The above equation satisfies these special cases whne written as

m x y = -( b^2 / a^2 ) x

So, the equation of the straight line through the midpoints of parallel chords is

(m)y + ( b^2 / a^2 ) x = 0.

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