Find the point to which the origin should be shifted after a translation of axes so that the following equations will have no first degree terms:
(i) y2 + x2 – 4x – 8y + 3 = 0
(ii) x2 + y2 – 5x + 2y – 5 = 0
(iii) x2 – 12x + 4 = 0
To find: The point to which origin has to be shifted such that there are no first-degree terms, i.e. there are no terms with (variable)1
We know that, when we transform origin from (0, 0) to an arbitrary point (p, q), the new coordinates for the point (x, y) becomes (x + p, y + q), and hence an equation with two variables x and y must be transformed accordingly replacing x with x + p, and y with y + q in original equation.
In following subproblems, we assume that origin has been shifted from (0, 0) to (p, q); therefore any arbitrary point (x, y) will also be converted as (x + p, y + q).
(i) y2 + x2 – 4x – 8y + 3 = 0
Substituting x and y with (x+p) and (y+q) respectively, we have
= (x+p)2 + (y + q)2 – 4(x + p) - 8(y + q) + 3 = 0
= x2 + p2 + 2px - y2 – q2 – 2qy – 4x – 4p – 8y – 8q + 3 = 0
= x2 + y2 + x(2p – 4) + y(2q – 8) + p2 + q2 – 4p – 8q + 3 = 0
For first degree term to be zero we have,
2p – 4 = 0 and 2q – 8 = 0
Giving us, p = 2 and q = 4.
Hence, the shifted point is (p, q) = (2, 4).
(ii) x2 + y2 – 5x + 2y – 5 = 0
Substituting x and y with (x+p) and (y+q) respectively, we have
= (x+p)2 + (y + q)2 – 5(x + p) + 2(y + q) - 5 = 0
= x2 + p2 + 2px - y2 – q2 – 2qy – 5x – 5p + 2y + 2q - 5 = 0
= x2 + y2 + x(2p – 5) + y(2q + 2) + p2 + q2 – 5p + 2q - 5 = 0
For first degree term to be zero we have,
2p – 5 = 0 and 2q + 2 = 0
Giving us, p = 5/2 and q = 1.
Hence, the shifted point is (p, q) = (5/2, 1).
(iii) x2 – 12x + 4 = 0
Substituting x and y with (x+p) and (y+q) respectively, we have
= (x+p)2 – 12(x + p) + 4 = 0
= x2 + p2 + 2px – 12x – 12p + 4 = 0
= x2 + x(2p – 12) + p2 – 12p + 4 = 0
For first degree term to be zero we have,
2p – 12 = 0.
Giving us, p = 2.
Hence, the shifted point is (p, q) = (2, q), where q can be any real number.