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" Prove that the area of the parallelogram formed by the tangents at the extremities of two conjugate diameters of an ellipse is constant, and is equal to 4 ab. "
A Treatise on the Analytical Geometry of the Point, Line, Circle, and Conic ... - Page 212
by John Casey - 1893 - 564 pages
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A Treatise on Plane Co-ordinate Geometry: With Numerous Examples ...

Isaac Todhunter - Geometry - 1855 - 332 pages
...'2 ' 172 • abab 3§. The excentric angles of two points P and Q are <f> and A' respectively; shew that the area of the parallelogram formed by the tangents at the extremities of the diameters through P and Q is —. — 7-77 -^ ; shew also that the area is sin (<p — <p) least...
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A Treatise on Plane Co-ordinate Geometry as Applied to the Straight Line and ...

Isaac Todhunter - Conic sections - 1858 - 334 pages
...sin4 a + &2 cos2 a. 38. The excentric angles of two points P and Q are <£ and <ft respectively; shew that the area of the parallelogram formed by the tangents at the extremities of the diameters through P and Q is — — rr, — -r-. ; shew also that the area is 0 sm (9 — <p)...
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The Mathematical Writings of Duncan Farquharson Gregory

Duncan Farquharson Gregory, Robert Leslie Ellis - Mathematics - 1865 - 320 pages
...applications of the theory. MATHEMATICAL NOTE.' THE area of the parallelogram formed by tangents applied at the extremities of any two conjugate diameters of an ellipse is constant. Let x, y be the coordinates of the extremity of one diameter, and x\ y those of the other ; 0, & the...
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Mathematical Questions with Their Solutions: From the ..., Volumes 4-6

1866 - 390 pages
...the semi-axes of an ellipse, and <j>, </>'the eccentric angles of two points P, Q on the curve; prove that the area of the parallelogram formed by the tangents at the ends of the diameters through P and Q is 4a6 ccsec (</>' — <J>). Solution by J. McDoWEiL, HA ; SW...
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An elementary manual of coordinate geometry and conic sections

James White - Conic sections - 1878 - 160 pages
...their extremities, is constant. The triangle TC T' = ia'. V sin TCT= ab = io6. From this it evidently follows that the area of the parallelogram formed by the tangents at the extremities of a pair of conjugate diameters, being equal to eight times the triangle, is equal to 4<z6. It is evident...
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Elements of the Conic Sections

Rufus B. Howland - Conic sections - 1887 - 76 pages
...: OE : F'D ; hence FG : I'L : : OE : F'D, or PL • OE = PL • PK= FG • F'D = FC2. THEOREM XXVI. The area of the parallelogram formed by the tangents at the extremities of two conjugate diameters of an ellipse is constant, and equals the rectangle on the axes. Let PC be...
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An Elementary Treatise on the Geometry of Conics

Sir Asutosh Mookerjee - Conic sections - 1893 - 197 pages
...and Note on Tangent- Proper lies, Ex. I., 1.] PR . PR' : QR . QR'=OP* : OQ' 2 . * PEOPOSITION XXXV. The area of the parallelogram formed by the tangents at the extremities of a pair of conjugate diameters is constant. (GD.PF=GA.CB.) The tangents at the extremities of two conjugate...
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The Elements of Coordinate Geometry

Sidney Luxton Loney - Coordinates - 1896 - 447 pages
...similar ellipse. 22. The eccentric angles of two points P and Q on the ellipse are 01 and 02 ; prove that the area of the parallelogram formed by the tangents at the ends of the diameters through P and Q is and hence that it is least when P and Q are at the end of...
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Examination Papers: A Supplement to the University Calendar

1903 - 788 pages
...axis, a point on the curve and the normal at that point, construct the focus and directrix. 4. Show that the area of the parallelogram formed by the tangents at the ends of two conjugate diameters of an ellipse is constant. If two conjugate diameters are given in...
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An Elementary Treatise on Graphs

George Alexander Gibson - Algebra - 1904 - 204 pages
...the coordinates of F are ~ a2sin20 + &2cos20 ' and that CF= J(a* +/) = ^. 24. Show from example 23 that the area of the parallelogram formed by the tangents at the ends of two conjugate diameters PCP, DCLf is constant, and equal to 4a6 or A A'. BB-, the rectangle...
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