Rotation of Axes (Change of Direction)
1) When the axes rotated through an angle 60, then new co-ordinates of three points are the following (3, 4).
Solution: Given that,
New co-ordinates are (3, 4)
X = 3, Y = 4
x = X cos θ – Y sin θ
= 3 cos60 – 4 sin60
= 3 (½) – 4 \(\frac{\sqrt{3}}{2}\),
= \(\frac{3-4\sqrt{3}}{2}\).
y = X sin θ + Y cos θ
= 3 sin60 + 4 cos60
= 3 \(\frac{\sqrt{3}}{2}\) + 4 (½)
= \(\frac{4+\sqrt{3}}{2}\).
Co-ordinate of P are \(\left( \frac{3-4\sqrt{3}}{2},\frac{4+\sqrt{3}}{2} \right)\).
2) Find the angle through which the axes are to be rotated so as to remove the xy term in the equation x² + 4xy + y² – 2x + 2y – 6 = 0.
Solution: Comparing the equation
x² + 4xy + y² – 2x + 2y – 6 = 0 with ax² + 2hxy + by² + 2gx + 2fy + c = 0
a = 1, h = 2, b = 1, g = -1, f = 1, c = -6
let θ be the angle of rotation of axes, then θ = ½ tan⁻¹\(\left( \frac{2h}{a-b} \right)\),
= ½ tan⁻¹ (4/ 1 – 1) = ½ tan⁻¹ (4/0)
= ½ tan⁻¹(∞) = ½ x π/2
θ = π/4
3) When the axes are rotated through an angle 45, the transformed equation of a curve is 17x² – 16xy + 17y² = 255. Find the original equation of the curve?
Solution: Angle of rotation = θ = 45
X = x cosθ + y sinθ
= x cos45 + y sin45 = (x + y)/2
Y = -x sinθ + y cosθ
= -xsin45 + ycos45 = (-x + y)/2
The original equation is 17x² – 16xy + 17y² = 255
⇒ \({{17\left( \frac{x+y}{2} \right)}^{2}}-16\left( \frac{x+y}{2} \right)\left( \frac{-x+y}{2} \right)+17{{\left( \frac{-x+y}{2} \right)}^{2}}=225\),
⇒ \(17\left( \frac{{{x}^{2}}+{{y}^{2}}+2xy}{2} \right)-16\left( \frac{{{x}^{2}}-{{y}^{2}}}{2} \right)+17\left( \frac{{{x}^{2}}+{{y}^{2}}+2xy}{2} \right)=225\),
⇒ 17 (x² + y²) – 8 (x² – y²) = 225
⇒ 19 x² + 25 y² = 225.