Weekly problem papers, with notes. [With] Solutions1885 |
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Side ix
... fixed . ' 7 , in the last factor , for + u2 read a2 . VIII . 99 99 XII . " " 2 , in first line , for a read " . دو ... point of falling over . ' = - read x = + . XXIX . 19 99 2 , for 1 - c2 read - c2 . X 14 " " 99 XXXIX . وو XLI ...
... fixed . ' 7 , in the last factor , for + u2 read a2 . VIII . 99 99 XII . " " 2 , in first line , for a read " . دو ... point of falling over . ' = - read x = + . XXIX . 19 99 2 , for 1 - c2 read - c2 . X 14 " " 99 XXXIX . وو XLI ...
Side 22
... point p evidently lies on the directrix , and T and P are equidistant from the directrix . 7. Let 2a be the angle ... fixed point ( § , n ) is ( x − h ) ( § - h ) + yn = h2 sin 2a , or h2 cos 2a - h ( x + § ) + x + yn = 0 . To find the ...
... point p evidently lies on the directrix , and T and P are equidistant from the directrix . 7. Let 2a be the angle ... fixed point ( § , n ) is ( x − h ) ( § - h ) + yn = h2 sin 2a , or h2 cos 2a - h ( x + § ) + x + yn = 0 . To find the ...
Side 23
... fixed points , and OA a fixed straight line . Then if any point P be taken on OA , and PB be joined , and a point R be taken on PB such that PR . PB : PO2 in a constant ratio , as P moves along OA it is required to find the envelope of ...
... fixed points , and OA a fixed straight line . Then if any point P be taken on OA , and PB be joined , and a point R be taken on PB such that PR . PB : PO2 in a constant ratio , as P moves along OA it is required to find the envelope of ...
Side 61
... fixed point on the circumference of a circle , and AB a chord which subtends half a right angle at S. Then it is clear that AB always touches a circle concentric with the given circle , and that the tangents to it from S are at right ...
... fixed point on the circumference of a circle , and AB a chord which subtends half a right angle at S. Then it is clear that AB always touches a circle concentric with the given circle , and that the tangents to it from S are at right ...
Side 81
... fixed point . 4 ― - " n being even . - By symmetry L lies on the diameter polar of A , which is a fixed straight 6. By the same method which was employed in XXXVI . No. 6 , we can shew that when a triangle inscribed in an ellipse has ...
... fixed point . 4 ― - " n being even . - By symmetry L lies on the diameter polar of A , which is a fixed straight 6. By the same method which was employed in XXXVI . No. 6 , we can shew that when a triangle inscribed in an ellipse has ...
Vanlige uttrykk og setninger
a+b+c ABCD axis bisects the angle Cambridge centre chord circle round circle will go circum-circle Clifton College coef College conic const coordinates cos² cosec Crown 8vo denote diameter directrix draw Edition ellipse envelope equal equation Fcap fixed point geometrical given expression hyperbola inscribed circle intersect Join Let ABC Let the tangent locus Mathematics meet middle point Monday afternoon morning nine-point circle nth term orthocentre PAPER parabola parallel perpendicular plane Professor quadrilateral radical axis radius right angles shew shewn sides Similarly sin² sin³ straight line tan² tangent Todh triangle Trinity College Tripos velocity vertex vertical
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