132 A \ ALTTIC GEOMETi: Y [Cii. \ I 



EXAMPLES ON CHAPTER VI 



1. Find the equation of the locus of 2zy-7jr + 4y = referred to 

 parallel axes through the point (~2, {). 



2. Transform the equation jr-4ry + 4y-6jr + 12y = to new 

 rectangular axes making an angle tair 1 J with the given axes. 



3. Transform y 1 xy 5z + 5y = to parallel axes through the 

 point (~5, ~5). Draw an appropriate figure. 



4. Transform the equation of example 3 to axes bisecting the angles 

 between the old axes. Trace the locus. 



5. To what point must the origin be moved (the new axes being 

 parallel to the old) in order that the new equation of the locus 



a _ 5xy _ 3 y _ 2z + 13 y - 12 = 

 shall have no terms of first degree ? 



SOLUTION. Let the new origin be (A, k) ; then x = xf -f h, y = \f -f Jt, 

 and the new equation is 



2(z f + A)_5(x' + ;0(y' + *)-3(y' + *)-2(;r' + h) + 13(y + *) -12 = 0, 

 *>., 2** - 5*y - 3y" -I- (4 h - 5k - 2)*' - (5h + Qk - 13)^ 



+ 2A - 5Ai - 3*' - 2A + 13fc - 12 = 0; 



but it is required that the coefficients of x 1 and y 1 shall be 0; i.e., h and 

 k are to be determined so that 



4A-5fc- 2 = 0, 



and 5A + 6*-13 = 0; 



hence h = V and /: = f . 



Therefore the new origin must be at the point (V. f)> and the new 

 equation is 



6. The new axes being parallel to the old, determine the new origin 

 so that the new equation of the locus 



a* - Zxy + y 2 + 10 x - lOy + 21 = 

 shall have no terms of first degree. 



7. Transform the equations x -H y - 3 = and 2*-3y-f4 = to 

 parallel axes having the point of intersection of these lines as origin. 



8. Transform the equation - + ^ = 1 to new rectangular axes through 

 the point (2, 3), and making the angle tan l ( \) with the old axes. 



9. Through what angle must the axes be turned that the new equa- 

 tion of the line 6z + 4/-2l = shall have no y-term ? Show this 

 geometrically, from a figure. 



