Transformation of Axes
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Direct answer
Coordinates are a bookkeeping choice: shifting the origin to (h, k) rewrites every point as x = X + h, y = Y + k (so the new coordinate is X = x − h), while a rotation of axes through angle α substitutes x = X cos α − Y sin α and y = X sin α + Y cos α. Shifting is chosen to kill first-degree terms — completing the square in coordinate language; rotation kills the xy cross term, its eliminating angle satisfying tan 2α = 2h/(a − b) for ax^2 + 2hxy + by^2 + ... = 0. Two combinations survive rotation unchanged — a + b and h^2 − ab — the built-in checks.
What you must remember
- Shift substitution: origin to (h, k) means replace x by X + h and y by Y + k in the old equation; the reversed sign (X − h) is the most common slip in the chapter.
- Rotation substitution: x = X cos α − Y sin α, y = X sin α + Y cos α, with α measured from the old x-axis toward the old y-axis.
- Killing the xy term: tan 2α = 2h/(a − b); when a = b the right side is infinite, 2α = 90° and α = 45° — exactly the xy = c^2 situation.
- Rotation invariants: a + b (the trace) and h^2 − ab (the discriminant) emerge from every rotation unchanged — verify answers with them.
- Translation behaviour: the quadratic-part coefficients never change under a shift; only the linear and constant terms move.
- Order of operations: shift first to remove linear terms (find the centre), then rotate to remove the cross term.
- Purpose: every transformation aims to restore a standard form — circle, parabola, ellipse, hyperbola — whose parameters are then read directly.
Rotating xy = 8 into standard form
The equation xy = 8 has a = b = 0 and h = 1/2, so tan 2α = 2h/(a − b) is undefined and 2α = 90°, giving α = 45°. Substitute x = (X − Y)/√2 and y = (X + Y)/√2: the product becomes (X^2 − Y^2)/2, so the equation reads (X^2 − Y^2)/2 = 8, or X^2 − Y^2 = 16. In standard form, X^2/16 − Y^2/16 = 1: a rectangular hyperbola with a = b = 4, eccentricity √2, and asymptotes X = ±Y — which are nothing but the original coordinate axes, now confirmed as the curve's asymptotes. The invariant check closes the loop on the un-scaled form: before rotation, a = 0, b = 0, h = 1/2 gives a + b = 0 and h^2 − ab = 1/4; after substitution, (X^2 − Y^2)/2 = 8 has a = 1/2, b = −1/2, h = 0, so a + b = 0 and h^2 − ab = 0 − (1/2)(−1/2) = 1/4. Both survive — but only before clearing the factor of 2; rescaling the equation rescales the invariants, so run the check on the substituted form as it first appears.
Where marks leak
JEE Main rarely announces "transform the axes"; it asks for centres, standard forms and classifications that require the machinery quietly. JEE Advanced asks it openly — rotate to remove the xy term, or identify the curve after a stated rotation. Three losses dominate: substituting x = X − h instead of X + h (test one concrete point, such as the old origin, to fix the direction); choosing the rotation angle with the wrong sign convention, which merely relabels the new axes; and skipping the invariant check that would have caught an arithmetic slip in the substitution. Keep the habit of verifying a + b and h^2 − ab before and after — thirty seconds that certify a page of expansion.
Frequently asked questions
What substitution shifts the origin to (h, k)?
x = X + h and y = Y + k, so that the new coordinates of the old origin are (−h, −k).
What angle removes the xy term from ax^2 + 2hxy + by^2?
The rotation through α with tan 2α = 2h/(a − b); if a = b, use α = 45°.
What quantities are invariant under rotation of axes?
The trace a + b and the discriminant h^2 − ab; both survive unchanged in every rotated frame.
What curve does xy = 8 become after a 45-degree rotation?
X^2 − Y^2 = 16, a rectangular hyperbola with the original axes as asymptotes.
Why shift before rotating?
Shifting first removes the linear terms and centres the curve, so the rotation then acts on a clean quadratic part — reversing the order doubles the algebra.