The Elements of Coordinate Geometry

Cover The Elements of Coordinate Geometry
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Find the equation to the chord joining the points on the circle r — 2a cos 6 whose vectorial angles are d^ and 6^, and deduce the equation to the tangent at the point 6^.
The equation to any straight line in polar coordinates is (Art. 88) ^ = rcos((9-a) (1).
If this pass through the points (2a; cos ^j^, 6^ and (2asin^2» ^2)' ^^^ have 2a cos d-^ cos (^j-a)=^ = 2acos ^.3 cos {9.^~a) (2).
Hence cos (2^^ - a) + cos a = cos (2^., - a) + cos a, i.e. 2^1 -a= -(2^0 -a), since d^ and ^2 ^^^ iiot» i'^ ge
...neral, equal.
Hence a — d-^ + d.^^ and then, from (2), j? = 2a cos 6-^ cos d.-^.
On substitution in (1), the equation to the required chord is r cos (^ - ^1 - ^2) = 2a cos ^1 cos ^2 (3).
The equation to the tangent at the point d^ is found, as in Art. 150, by putting 6^ = 6^ in equation (3).
We thus obtain as the equation to the tangent rcos (^-2^^) — 2acos2^j^. • As in the foregoing article it could be shewn that the equation to the chord joining the points d^ and 6^ on the circle r= 2a cos {d - 7) is r cos {d -9-^- 6., + 7] — 2a cos {9-^ - 7) cos (9.^ - 7) and hence that the equation to the tangent at the point ^^ is r cos (^ - 2^j + 7) = 2a cos2 (^j - 7).


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