Chords bisected by a point?

The equation S can be equivalently written as

(x - 4)^2 / 70 + (y + 20)^2 / 1470 = 1

To simplify we move the center to the origin (call the equation E),

E: x^2 / 70 + y^2 / 1470 = 1

We move the point P[ 7, - 29 ] the same amount (call it Q):

Q[ 3, - 9 ]

Using the parametrization of ellipse, a point on E (call it R) is expressed as

R[ √70 (h^2 - 1) / (h^2 + 1), 14√30 h / (h^2 + 1) ]

If we call another point U, RU must be bisected by Q, so the coordinates of U are

U [ 3*2 - √70 ( h^2 - 1 ) / ( h^2 + 1 ), - 9*2 - 14√30 h / ( h^2 + 1 ) ]

= [ ( ( 6 - √70) h^2 + ( 6 + √70 ) ) / ( h^2 + 1) , ( - 18h^2 - 14√30 h - 18 ) / ( h^2 + 1 ) ]

U must be on E, so

21 { ( ( 6 - √70) h^2 + ( 6 + √70 ) ) / ( h^2 + 1) }^2 + { ( - 18h^2 - 14√30 h - 18 ) / ( h^2 + 1 ) }^2 = 1470

http://www.wolframalpha.com/input/?i=21+%28+%28+%2...

h = ( 20√10 + 7√30 ) / ( 7√70 - 30 )

or h = ( - 20√10 + 7√30 ) / ( 7√70 - 30 )

When h = ( 20√10 + 7√30 ) / ( 7√70 - 30 ),

R = [ 3 + 2√(30/7), 2√210 - 9 ], U = [ 3 - 2√(30/7), - 2√210 - 9 ]

When h = ( - 20√10 + 7√30 ) / ( 7√70 - 30 ),

R = [ 3 - 2sqrt(30/7), - 2 sqrt(210)-9 ], U = [ 3 + 2sqrt(30/7), 2 sqrt(210)-9 ]

Equation for line RU is

http://www4a.wolframalpha.com/input/?i=line+throug...

RU: - 7x + y + 30 = 0

We translated P and E at the start, so we translate RU back. Finally, equation of our chord of S that is bisected by P is

- 7( x - 4 ) + ( y + 20 ) + 30 = 0

- 7x + y + 78 = 0

***

The curve is an ellipse, so the number of correct answers will be limited.

Really, you have 2 equations to work with. The first one is the one you described. The 2nd is the midpoint formula. If (x1, y1) and (x2,y2) are the points on the ellipse, then

SQRT((x1-7)^2 + (y1+29)^2) = SQRT(x2-7)^2 + (y2+29)^2.

I'd work from there.

I could only find one so far (still hunting):

y + 29 = 7.008055(x – 7)

Cheers! ☺

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