Transformation of coordinates involves changing the coordinate system (the axes and/or origin) to simplify the equation of a curve.
This transformation involves shifting the origin (0, 0) to a new point (h, k) without changing the direction (orientation) of the axes. The new axes remain parallel to the old axes.
Let (x, y) be the coordinates of a point P with respect to the old axes (OX, OY), and let (x', y') be the coordinates of the same point P with respect to the new axes (O'X', O'Y').
The transformation formulas are:
x = x' + h
y = y' + k
Or, rearranging for the new coordinates:
x' = x - h
y' = y - k
Question: Transform the equation x² + y² - 4x + 6y - 3 = 0 by shifting the origin to the point (2, -3).
Solution:
Here, h = 2 and k = -3.
Substitute x = x' + 2 and y = y' - 3 into the equation:
(x' + 2)² + (y' - 3)² - 4(x' + 2) + 6(y' - 3) - 3 = 0
(x'² + 4x' + 4) + (y'² - 6y' + 9) - 4x' - 8 + 6y' - 18 - 3 = 0
Now, group the terms:
x'² + y'² + (4x' - 4x') + (-6y' + 6y') + (4 + 9 - 8 - 18 - 3) = 0x'² + y'² - 16 = 0x'² + y'² = 16
The new, simplified equation is a circle centered at the new origin.
This transformation involves rotating the axes (OX, OY) by an angle θ about the origin, without changing the origin itself. This is an **orthogonal transformation** because it preserves distances and angles.
Let (x, y) be the coordinates of a point P w.r.t. the old axes and (x', y') be the coordinates w.r.t. the new axes.
The transformation formulas are:
x = x' cos(θ) - y' sin(θ)
y = x' sin(θ) + y' cos(θ)
These can be remembered using the following table:
| x' | y' | |
|---|---|---|
| x | cos(θ) | -sin(θ) |
| y | sin(θ) | cos(θ) |
Invariants are quantities in an equation that do not change their value after an orthogonal transformation (like rotation). This is a very useful concept for identifying conics.
Consider the general equation of the second degree: ax² + 2hxy + by² + 2gx + 2fy + c = 0.
After rotating the axes by an angle θ, the equation becomes a'x'² + 2h'x'y' + b'y'² + 2g'x' + 2f'y' + c' = 0.
The following quantities are **invariants**:
c = c' (Note: The syllabus mentions orthogonal transformation, which is typically rotation. Translation *will* change the constant term).a + b = a' + b'ab - h² = a'b' - h'²a + b and ab - h². Remember these two, as they help determine the nature of the conic.
A second-degree equation can sometimes represent two straight lines. We study two cases.
A homogeneous equation of the second degree is of the form:
ax² + 2hxy + by² = 0This equation *always* represents a pair of straight lines passing through the **origin** (0, 0).
The two lines can be found by factoring the equation. If we divide by x², we get a quadratic in (y/x): b(y/x)² + 2h(y/x) + a = 0.
If m₁ and m₂ are the roots (slopes) of this equation, the two lines are y = m₁x and y = m₂x.
The angle θ between the two lines ax² + 2hxy + by² = 0 is given by the formula:
tan(θ) = | 2 * sqrt(h² - ab) / (a + b) |
From this formula, we can determine the nature of the lines:
h² > abh² = abh² < abSpecial Cases:
Condition: a + b = 0
(i.e., Coefficient of x² + Coefficient of y² = 0)
Condition:h² - ab = 0orh² = ab
The full general equation is:
S = ax² + 2hxy + by² + 2gx + 2fy + c = 0This equation represents a pair of straight lines (which may be parallel) if and only if a specific condition is met. This condition can be expressed in two ways:
Condition 1 (The "long" formula):
abc + 2fgh - af² - bg² - ch² = 0
Condition 2 (The determinant form):
The determinant Δ (delta) must be zero.
Δ =Δ = a(bc - f²) - h(hc - fg) + g(hf - bg) = 0
a h g h b f g f c
Given the pair of lines ax² + 2hxy + by² = 0, the combined equation of the two lines that bisect the angles between them is:
(x² - y²) / (a - b) = xy / h
If a = b, the equation becomes x² - y² = 0, or (x - y)(x + y) = 0, meaning the bisectors are y = x and y = -x.
If h = 0, the equation becomes xy = 0, meaning the bisectors are the axes x = 0 and y = 0.
Important Note: The two angle bisectors are always perpendicular to each other. We can check this using the condition A + B = 0.
Rearranging the bisector equation: h(x² - y²) = (a - b)xy
hx² - (a - b)xy - hy² = 0
Here, A = h and B = -h.
A + B = h + (-h) = 0. Thus, they are always perpendicular.
x = x' + h, y = y' + k. Used to remove first-degree terms.x = x'cos(θ) - y'sin(θ), y = x'sin(θ) + y'cos(θ). Used to remove the xy term.a + b and ab - h² are constant.tan(θ) = |2*sqrt(h² - ab) / (a + b)|a + b = 0h² = ababc + 2fgh - af² - bg² - ch² = 0. This is a MUST-KNOW formula.(x² - y²) / (a - b) = xy / h. This is also a MUST-KNOW formula.